Indefinite Integral. Chapter Integration - reverse of differentiation

Size: px
Start display at page:

Download "Indefinite Integral. Chapter Integration - reverse of differentiation"

Transcription

1 Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the speed of the process t tht moment.. Integrtion - reverse of differentition Consider the function F (x) x. Suppose we write its derivtive s f(x), tht is f(x) df. It is esy to see tht f(x) 6x. This process is illustrted in the following figure. Suppose now tht we try to differentite F (x) x + 6 insted. Clerly, the nswer to this question is gin the function 6x. We sy tht F (x) x or F (x) x + 6 is the ntiderivtive of f(x) 6x. We cn see tht n ntiderivtive is not uniquely determined becuse the ntiderivtives of 6x cn lso be x +, x π nd ny function of the form x + C, where C is n rbitrry constnt. The reson why ll of these functions hve the sme derivtive is tht the constnt term disppered fter differentition. So, more generlly, if F (x) is n ntiderivtive of f(x), then the ntiderivtive of f(x) is lso every function F (x) + C, where C is whtever constnt. Exmple.. () Let F (x) e x + cos x, find df.

2 CHAPTER. INDEFINITE INTEGRAL (b) Write down severl ntiderivtives of f(x) e x sin x. () df ex sin x e x sin x. (b) We cn deduce from () tht n ntiderivtive of e x sin x is e x + cos x. All other ntiderivtives of f(x) e x sin x will tke the form F (x) + C where C is constnt. So, the following cn ll be the ntiderivtives of f(x): e x + cos x 7, e x + cos x 5, e x + cos x, e x + cos x +. From these exmples, we hve the following definition.. Indefinite Integrl If the function F (x) is n ntiderivtive of f(x), then the expression F (x) + C, where C is n rbitrry constnt, is clled the indefinite integrl of f(x) with respect to x nd is denoted by f(x), i.e., The function f(x) is clled the integrnd, the constnt of integrtion. Exmple.. Find x. f(x) F (x) + C. the integrl sign, x is clled the vrible of integrtion nd C Let y x, then dy x. Hence, we verify tht x x. In generl, dding ny constnt C to x gives the sme derivtive. Hence, the integrl of the function y x is given by x x + C. Exmple.. Find Since Hence x. d ln x x, x ln x + C.

3 CHAPTER. INDEFINITE INTEGRAL Exmple.4. Let n be n integer not equl to. Find x n. Since Hence d xn+ (n + )x n. (n + )x n x n+ + C, dividing (n + ) nd dding n rbitrry constnt gives x n n + xn+ + C. Exmple.5. Find cos x. Since Hence d sin x cos x, cos x sin x + C... Bsic Properties of Indefinite Integrls Since integrtion is the reverse process of differentition, we cn derive some bsic properties of indefinite integrls from differentition.. kf(x) k f(x), where k is constnt.. f(x) ± g(x) f(x) ± g(x) Proof. Suppose F (x) is the ntiderivtive of f(x), i.e., d F (x) f(x). d d [kf (x)] k F (x) kf(x) By the definition of indefinite integrl, kf(x) kf (x) + C. On the other hnd, k f(x) k [F (x) + C] kf (x) + C. Since C nd C re rbitry constnts, kf(x) k f(x).

4 CHAPTER. INDEFINITE INTEGRAL 4 Proof. Suppose F (x) nd G(x) re the ntiderivtive of f(x) nd g(x) respectively. i.e., nd d G(x) g(x). d [F (x) ± G(x)] d F (x) ± d G(x) f(x) ± g(x) By the definition of indefinite integrl, [f(x) ± g(x)] F (x) ± G(x) + C d F (x) f(x) F (x) + C ± G(x) + C f(x) ± g(x)... Tble of bsic Integrls Similr to our exmples. to.5, by reversing our lists of derivtives nd rules, we will hve list of elementry integrls results s shown in the following tble : Exmple.6. Find (x x ). (x ) x ( x x ) x x + C

5 CHAPTER. INDEFINITE INTEGRAL 5 Exmple.7. x 6x + C x + 6 Find x. x + 6 x ( + 6 x ) ( + 6x ) x 6 x + C Exmple.8. Find sec x(cos x + tn x). sec x(cos x + tn x) (sec x cos x + sec x tn x) ( + sec x tn x) x + sec x + C Exmple.9. Find (6x + x ex ). (6x + ( ) x x ex ) 6 + ln x e x + C x + ln x e x + C Exmple.. Find sin x. sin x cos x sec x tn x + C

6 CHAPTER. INDEFINITE INTEGRAL 6 Exmple.. + cos θ Find cos θ dθ + cos θ + cos θ cos θ dθ cos θ + cos θ + cos θ dθ + cos θ + cos θ sin dθ θ (csc θ + cot θ csc θ + cot θ) dθ cot θ csc θ + (csc θ ) dθ cot θ csc θ θ + C. Integrtion by Substitution x Some indefinite integrls like + cnnot be found directly by using the bsic integrtion tble. In this cse, we my trnsform the indefinite integrl into nother indefinite integrl tht we know how to integrte. This method is known s integrtion by substitution. In generl, we hve: If F (u) is the ntiderivtive of f(u), i.e. F (u) f(u), nd u g(x) is differentible function of x, then we hve f[g(x)]g (x) f(u)du F (u) + C. Proof. Since F (u) is n ntiderivtive of f(u). We hve f(u) du F (u) + C. By the chin rule, d df (u) F (u) du du f(u) g (x) f [g(x)] g (x). By the definition of indefinite integrl, f[g(x)]g (x) F (u) + C. Hence, Exmple.. Find x x +. f[g(x)]g (x) f(u)du F (u) + C.

7 CHAPTER. INDEFINITE INTEGRAL 7 Let u x +. Then du x x + (u )( u) du (u u ) du 5 u 5 u + C 5 (x + ) 5 (x + ) + C Exmple.. Find x 5 x 5x +. Let u x 5x +. Then du (x 5) x 5 x 5x + u du u + C x 5x + + C Exmple.4. Find e x 4. Let u x 4. Then, du or 4 du 4 e x 4 e u ( 4) du 4e u + C 4e x 4 + C

8 CHAPTER. INDEFINITE INTEGRAL 8 Exmple.5. x ln x Find. x Let u ln x. Then, du x x ln x x (x ln x x ) x u du ( ) x u + C x (ln x) + C Exmple.6. Find x csc ( x ). Let u x. Then, du x or du x x csc ( x ) csc u du ( cot u) + C cot( x ) + C Exmple.7. Find cos θ (sin θ + ) dθ. Let u sin θ +. Then, du cos θ dθ cos θ (sin θ + ) dθ u du u + C sin θ + + C.. Integrtion involving x, x nd + x In mny cses, the shortest method of integrting such expressions: x, x nd + x where >, is to chnge the vrible s follows:

9 CHAPTER. INDEFINITE INTEGRAL 9. For integrnds involving x. We use the substitution x sin θ, π θ π.. For integrnds involving x. We use the substitution x sec θ, θ π, θ π.. For integrnds involving + x. We use the substitution x tn θ, π < θ < π. In ech of the bove cses, the rdicl sign in the integrnd will be eliminted. Exmple.8. Find 4 x. Let x sin θ. Then, cos θ dθ. 4 x cos θ dθ cos θ θ + C sin x + C Exmple.9. x 9 Find. x Let x sec θ. Then, sec θ tn θ dθ. x 9 x tn θ ( sec θ tn θ) dθ 9 sec θ tn θ dθ (sec θ ) dθ tn θ θ + C x 9 cos x + C (By drwing right-ngled tringle nd using pythgoren theorem) Exmple.. Find x + x. let x tn θ. Then, sec θ dθ. x + x tn θ sec θ (sec θ) dθ

10 CHAPTER. INDEFINITE INTEGRAL tn θ dθ tn θ θ + C x tn x + C.4 Integrtion by Prts If u(x) nd v(x) re differentible functions of x, then u(x)d(v(x)) u(x)v(x) v(x)d(u(x)). Proof. Since [u(x)v(x)] u(x)v (x) + v(x)u (x). Hence u(x)v (x) [u(x)v(x)] v(x)u (x). Then, i.e., u(x)v (x) [u(x)v(x)] u(x)d(v(x)) u(x)v(x) v(x)d(u(x)) v(x)u (x). Some integrls such s ln x, x cos x nd e x sin x cnnot be found by the techniques we hve lernt before. We shll use the bove method (integrtion by prts) to find these types of integrls. Exmple.. Find x 4 ln x. x 4 ln x ln x d( x5 5 ) (ln x)( x5 x 5 5 ) d(ln x) 5 5 x5 ln x x x5 ln x x5 5 + C

11 CHAPTER. INDEFINITE INTEGRAL Exmple.. Find x e x. x e x x d(e x ) x e x e x d(x ) x e x xe x x e x x d(e x ) x e x xe x + e x x e x xe x + e x + C Exmple.. Find x sin x. x sin x x d(cos x) x cos x + cos x d(x ) x cos x + x cos x x cos x + x d(sin x) x cos x + x sin x sin x x cos x + x sin x + cos x + C 4 Exmple.4. Find e x sin x. e x sin x e x d( cos x) e x cos x + e x cos x e x cos x + e x d(sin x) [ e x cos x + e x sin x ] e x sin x

12 CHAPTER. INDEFINITE INTEGRAL e x cos x + e x sin x 4 e x sin x Hence, 5 e x sin x e x cos x + e x sin x + C e x sin x 5 ( ex cos x + e x sin x) + C Exercises. Find (x )(x + )( x). e x x. Find.. Find (sec θ + csc θ)(sec θ csc θ) dθ. 4. Find cos x tn x cos x. 5. Find (cos x + sin x cot x). csc 4 x 6. Find cot. x x ln x 7. Find. x 8. Find cos x sin x. 9. Find x e x.. Find e x ex +.. Find cos x sin 4 x.. Find tn θ sec 6 θ dθ.. Find 4. Find 5. Find 6. Find. (9 x ) (x + ) x. x 4x. x 5 x x +.

13 CHAPTER. INDEFINITE INTEGRAL 7. Find x + x Find x sin x sin 4x. 9. Find sin(ln x).. Find e x sin x.. () Find x +. (b) Using the substitution u e x nd the result of (),find (c) Using the result of (), find x + x +.. () Find tn 6 x. e x + e x. (x ) 5 (b) Using the result of (), find. x. () Show tht d tn x + cos x. (b) Using (), or otherwise, find x + sin x + cos x. 4. () Prove tht sec x ln sec x + tn x + C (b) Using the result of (), evlute sec x nd sec x.

14 Chpter Definite Integrl We hve introduced ll the skills nd procedures of finding indefinite integrl. Now, we re ble to ntidifferentite wide rnge of elementry functions. We hve to mster ll the skills of indefinite integrl before we move on to this chpter. In this chpter, we del with definite integrl. The min difference between definite nd indefinite integrl is the existence of upper nd lower limits. Usully, we denote the upper limit by b nd lower limit by. The nottion of definite integrl is b f(x).. Definition of Definite Integrls For function f defined on [, b], prtition P of [, b] into collection of n subintervls [x, x ], [x, x ],..., [x n, x n ], nd for ech k,,..., n, point ξ k in [x k, x k ], the sum under grph is given by or in summtion nottion f(ξ )(x x ) + f(ξ )(x x ) f(ξ n )(x n x n ) n f(ξ k )(x k x k ) k n f(ξ k ) x. Usully, we will tke P to be n regulr prtition (i.e. x b ). This is clled Riemnn sum of the n function f. Notice tht s n tends to infinity, the sum would be more nd more ccurte representtion of the re under grph. The definite integrl of f from to b is defined to be b f(x) lim n n k k f(ξ k )( b n ). 4

15 CHAPTER. DEFINITE INTEGRAL 5 Exmple.. Compute the integrl 4 x by computing Riemnn sums for regulr prtition. Notice tht, b 4 nd f(x) x. For regulr prtition nd for ech positive integer n, we hve x 4 n Then, 4 n ; x k + k x 4k, for k,,, n; n ( ) 4k f(ξ k ) n n f(ξ k ) x k 64k, for k,,, n. n n ( ) ( ) 64k 4 n n n 56k k k 56 n 4 n 4 n k k 56n (n + ) 4n 4. (Notice tht the lst step of the simplifiction follows from summtion formul.) Then, by the definition of definite integrl, 4 x lim n k lim n n f(ξ k ) x ( 64k n ) ( ) 4 n

16 CHAPTER. DEFINITE INTEGRAL 6 56n (n + ) lim n 4n Bsic properties of Definite Integrls In the lst section, we hve defined definite integrl of f(x) on the closed intervl [, b]. This is to imply tht < b. We shll lso define integrls in which the upper limit is less thn the lower limit. Definition. Let f(x) be integrble on [, b]. Then, the definite integrl of f(x) from b to is defined by b f(x) b Directly from this definition, we hve the following properties: b kf(x) k b b b [f(x) ± g(x)] f(x) f(x) c f(x), where k is constnt. b f(x) + f(x) ± c b b g(x) f(x), where < c < b. f(x). We will prove property. The proofs of the other properties re esy nd trivil. They re left to our students s n exercise. Proof. Since Then, f(x) f(x) f(x). f(x).

17 CHAPTER. DEFINITE INTEGRAL 7. Fundmentl Theorem of Clculus As you will see in section. tht it is quite difficult to evlute definite integrl directly from the definition. This problem cn be esily solved by mens of the following fmous theorem: If f(x) is continuous function on [, b] nd F (x) is the ntiderivtive of f(x) (i.e. F (x) f(x)). We hve, b f(x) [F (x)] b F (b) F (). The Fundmentl Theorem of Clculus enbles us to evlute definite integrls by finding the indefinite b integrls. To evlute the definite integrl f(x), we first find the indefinite integrl f(x). Then, we substitute the upper nd lower limits into the resulting function to evlute the vlue of b f(x). In other words, we will repet ll the skills of the lst chpter gin nd do the upper nd lower limits substitution to the finl function without the integrtion constnt C. Note: For Integrtion by Substitutions, we need to chnge the upper nd lower limits to the new upper nd lower limits for the new vrible fter substitution. Exmple.. Evlute (x 4x ). [ ] (x 4x ) x x 4 [ ] () 4 [ ] ( ) ( ) 4 Exmple.. Evlute π sec x tn x. π sec x tn x [ ] π sec x ( π ) sec sec

18 CHAPTER. DEFINITE INTEGRAL 8 Exmple.4. Evlute (x + 4x) (x + 4x) (x + 4x). (x + 4x) [ x (x + 4x) + (x + 4x) + x + () 7 ] (x + 4x) Exmple.5. Evlute 4 6x x + 9. Let u x + 9. Then, du x. When x, u 9 nd x 4, u 5. Exmple x x u du [ ] 5 u 9 [5 7] 96 Evlute e e ln x + x. Let u lnx +. Then, du x. When x e, u nd x e, u. e e ln x + x u du ] [ u 5

19 CHAPTER. DEFINITE INTEGRAL 9 Exmple.7. Evlute π 4 tn x sec x. Let u sec x. Then du sec x tn x. When x, u nd x π 4, u. π 4 tn x sec x (u ) du [ ] u u [ ( ) ] + [ ] () Exmple.8. Evlute x 9 x. Let x sin θ. Then, cos θ dθ. When x, u nd x, u π. x π 9 x ( sin θ) cos θ dθ cos θ π π sin θ dθ ( cos θ) d(cos θ) [ cos θ cos θ ] π 7 cos π cos π [ ] + 7 cos cos [ 7 ] 8 Exmple.9. Evlute xe x. xe x x d(e x )

20 CHAPTER. DEFINITE INTEGRAL Exmple.. [xe x ] e x [ ()e (e ) ] [e x ] e [ e e ] Evlute π x cos x. π x cos x π x d(sin x) [ x sin x ] π π x sin x π x d(cos x) π [x cos x] π cos x [π cos π] [sin x] π π [ ] π.4 Definite Integrls of Even, Odd nd Periodic Functions.4. Even nd Odd functions A function f(x) is n even function if f( x) f(x) for ll vlues of x. Exmple.. f(x) x is n even function since f( x) ( x) x f(x). A function f(x) is n odd function if f( x) f(x) for ll vlues of x. Exmple.. f(x) x is n odd function since f( x) ( x) x f(x). When f(x) is n even function or n odd function, we hve the following result. f(x), if f(x) is even. f(x), if f(x) is odd.

21 CHAPTER. DEFINITE INTEGRAL Proof. Consider f(x). f(x) f(x) + Let u x. Then, du. When x, u nd x, u. Hence, f(x) f(x) When f(x) is even function, f( x) f(x). f( u) ( du) f( u) du f( x) f( x) + f(x) When f(x) is even function, f( x) f(x). f(x) f(x) f(x) f(x) Exmple.. Evlute π 4 π 4 cos x. Let f(x) cos x. Notice tht f( x) cos( x) cos x f(x). Hence, cos x is n even function. π 4 π 4 cos x π 4 cos x [sin x] π 4 [sin π ] 4 sin Exmple.4. Evlute Let f(x) x x +. x x x. Notice tht f( x) + x ( x) f(x). Hence, + x + is n odd function. x x +

22 CHAPTER. DEFINITE INTEGRAL.4. Periodic Functions If function f(x) is periodic function with period T, then f(x + T ) f(x) for ll vlues of x. Exmple.5. f(x) sin x is periodic function with period π. Note: If f(x + T ) f(x), then f(x + nt ) f(x), where n is n integer. When f(x) is periodic function with period T, we hve the following result. Proof. nt T f(x) n f(x), where n is n integer. Cse I: n is positive integer. Consider Similrly, Hence, T T nt f(x) T T nt f(x) + f(x) + + f(x) T (n )T f(x). Let u x T. Then du. When x T, u nd x T, u T. nt T T T T (n )T nt f(x) f(x) f(x) f(x) T T T T. n Cse II: n. The sttement is trivil. Cse III: n is negtive integer. T T T f(u + T ) du f(u) du f(x) f(x) f(x) T f(x) + f(x) T nt f(x) + f(x) (n )T Let u x nt nd m n. Then, du. When x, u nt mt nd x nt, u nt f(x) mt mt f(u) du f(u) du

23 CHAPTER. DEFINITE INTEGRAL To conclude, nt T m n T f(u) du (by the result of cse I) f(x) T f(x) n f(x), where n is n integer. Exmple.6. It is given tht f(x) is periodic function with period π, () Show tht b+nπ (b) Hence, evlute () +nπ π f(x) f(x). b π f(x) 4 nd π f(x). f(x), where, b re rel numbers nd n is n integer. Let u x nπ. Then du. When x + nπ, u nd x b + nπ, u b. b+nπ +nπ f(x) b b b f(u + nπ) du f(u) du f(x) (b) π f(x) 4π+ π 4π (π) f(x) 4π+ π f(x) + 4π π f(x) + π f(x) + 4 () f(x) f(x)

24 CHAPTER. DEFINITE INTEGRAL 4 Exercises. Evlute the following definite integrls from the definition of definite integrls. () (b). Evlute. Evlute 4. Evlute 5. Evlute 6. Evlute 7. Evlute 8. Evlute (4x + ). (7 + x ). π 6 4 e 5 ( 9. () Evlute ) sin x + + sin x ( ) x + x. e x e x + e x x ln x + x. (x + )e x +x. x 6 x. (x + ) x.. e x e x. + e x (b) Using the substitution u x nd the result of (), evlute. Evlute. Evlute. Evlute. Evlute 4 π π 4. () Evlute x ln x. e x (sin x + cos x). x e x. x cos x. π x sec x. (b) Using the result of (), evlute 5. Evlute π π (e x + e x ) sin x. π x tn x. (x + )(x + ).

25 CHAPTER. DEFINITE INTEGRAL 5 6. Suppose f(x) is periodic function with period T. () Using the substitution u x + T, show tht numbers. (b) Using the result of (), show tht 7. Evlute 8. Evlute x x. x e x. T b f(x) 9. () Using the substitution u x, show tht (b) Using the result of (), evlute. () Show tht π f(x) π 4 f(x) +T x x + (x 4). f( π x). b+t +T (b) Using the result of (), evlute the following definite integrls. i. ii. π π cos x sin x sin x + cos x. sin x sin x + cos x. (Hint: + b ( + b)( b + b )) f(x), where nd b re rel f(x), where is rel number. x x + (x ) (x ) x + (x ).

26 Chpter 4 Appliction of Integrtion In this chpter, we will see how definite nd indefinite integrls be pplied to rel life sitution. The first ppliction of indefinite integrl would be on stright line motion nd the next ppliction would be on grphs nd functions. 4. Stright Line Motion Consider the motion of n object long stright line. If the displcement of the object from reference point t ny time t is s, then its velocity v t time t is given by v ds nd its ccelertion t time t is dt given by dv dt d s dt. Conversely, the displcement nd the velocity of the object cn be determined from the given velocity nd ccelertion respectively. For ny given ccelertion t time t, the velocity v nd displcement s cn be found by Exmple 4.. s v dt v dt. An object moves long stright line through fixed point O. The ccelertion ms of the object t seconds fter the object hs pssed through O is given by t. The object moves with velocity of 8 ms when t. () Find the velocity of the object when t 6. (b) Find the displcement of the object from O when t 6. () Let v ms be the velocity of the object. v dt (t ) dt 6

27 CHAPTER 4. APPLICATION OF INTEGRATION 7 5t t + C When t, v 8. Hence, C 8 5() + (). i.e. v 5t t +. When t 6, v 5(6) (6) + 7. (b) Let s m be the displcement of the object from O. s v dt (5t t + ) dt 5 t t + t + C When t, s. Hence, C. i.e. s 5 t t + t. When t 6, s 5 (6) (6) + (6) 6. Exmple 4.. A bll is thrown verticlly upwrds from the ground. Its velocity v ms fter t seconds is given by v t. Let s m be the height of the bll bove the ground. () Express s in terms of t. (b) Find the time tken for the bll to rech the highest point nd the mximum height reched. (c) Find the totl distnce trvelled by the bll from t to t. () s v dt ( t) dt t 5t + C When t, s. Hence, C. i.e. s t 5t. (b) When the bll reches the highest point, its velocity would be zero. i.e. v. Hence, t nd t. The time tken for the bll to rech the highest point is s. When t, s () 5().

28 CHAPTER 4. APPLICATION OF INTEGRATION 8 Hence, the mximum height reched is m. (c) When t, s () 5() 5. When t, s () 5() 5. From (b), the bll will rech its highest position nd strt to fll fter seconds. Hence, the totl distnce trvelled by the bll from t to t [( 5) + ( 5)] m 4. Geometricl Appliction We hve lernt tht the slope of curve y f(x) t ny point (x, y) is given by f (x) or dy. In prticulr, f (x ) is the slope of the curve t (x, y ). If the slope of curve t ny point is given, we cn use integrtion to find the eqution of the curve. Exmple 4.. The slope of curve t ny point (x, y) is given by dy (, ), find the eqution of the curve. x 4. If the curve psses through the point dy x 4 y (x 4) x 4x + C Since the curve psses through (, ), () 4() + C C Hence, the eqution of the curve is y x 4x +. Exmple 4.4. At ny point (x, y) of curve, d y x + 6. If (, 7) nd (, ) lie on the curve, find the eqution of the curve. d y x + 6 dy (x + 6) dy 6x + 6x + C

29 CHAPTER 4. APPLICATION OF INTEGRATION 9 Then, y (6x + 6x + C ). Hence, y x + x + C x + C. Since (, 7) nd (, ) lies on the curve, we hve 7 () + () + C () + C nd () + () + C () + C. On solving, C 5 nd C 7. i.e. The eqution of the curve is y x + x 5x 7. Exmple 4.5. At ny point (x, y) of curve, d y 4x. If the slope of the curve t (, 8) is 9, find the eqution of the curve. Since the slope of the curve t (, 8) is 9 Since (, 8) is point on the curve d y 4x dy (4x ) 8x x + C 9 8( ) ( ) + C C 7 dy 8x x + 7 y (8x x + 7) x 4 5x + 7x + C 8 ( ) 4 5( ) + 7( ) + C C i.e. The eqution of the curve is y x 4 5x + 7x +.

30 CHAPTER 4. APPLICATION OF INTEGRATION 4. Are of Plne Figure From the definition of integrls, we know tht b f(x) lim n n k f(ξ k )( b n ). Hence, the re of region bounded by curve f(x), the x-xis nd the line x nd x b cn be evluted by the definite integrls: b f(x). Similrly, the re of region bounded by curve f(y), the y-xis nd the line y nd y b cn be evluted by the definite integrls: Exmple 4.6. b f(y) dy. Find the re of the region bounded by the line y x +, the x-xis nd the lines x nd x. The required re: (x + ) [ x + x [ 4 ] ] 5 Exmple 4.7. Find the re of the region bounded by the curve y, the x-xis nd the lines x nd x. x Notice tht the region is below the x-xis when x lies between nd. Hence, the required re: ( x ) [ ] x Exmple 4.8. Find the re of the region bounded by the curve y x(x )(x ) nd the x-xis. When y, x, or. Hence, the x-intercepts of the curve re, nd. Notice tht from x to x, the re region is bove the x-xis while from x to x, the re region is below the x-xis. Hence, the required re: x(x )(x ) x(x )(x ) (x x + x) (x x + x)

31 CHAPTER 4. APPLICATION OF INTEGRATION [ x 4 4 x + x ] [ ] x 4 4 x + x Exmple 4.9. Find the re of the region below the x-xis bounded by the curve x + y 6 nd the line x. Notice tht y ± 6 x nd the required re is below the x-xis. Hence, we will tke y 6 x. The required re: 6 6 x 6 6 x Let x 6 sin θ. Then 6 cos θ dθ. When x, θ π 6 nd x 6, θ π. 6 6 x π 8 8 π 6 π 6 cos θ dθ π 6 ( + cos θ) dθ [θ + sin θ ] π π + 9 π 6 Exmple 4.. Find the re of the region bounded by the curve x y(y )(y + ) nd the y-xis. Notice tht we re now considering the re with respect to the y-xis insted of the x-xis. When x, y, or. Hence, the y-intercepts of the curve re, nd. Moreover, the required re is leftwrd of the y-xis from y to y nd rightwrd of the y-xis from y to y. The required re: y(y )(y + ) dy y(y )(y ) dy [ y 4 7 (y y y) dy 4 y y ] (y y y) dy [ y 4 4 y y ] Exmple 4.. Find the re of the region bounded by the curve y x nd the line y x. By solving y x nd y x simultneously, the x-coordinte of their points of intersection re x nd x. Notice tht when x, x x.

32 CHAPTER 4. APPLICATION OF INTEGRATION Hence, the required re: (x x ) ] [x x 4 Exmple 4.. Find the re of the region bounded by the curve y sin x+ nd y sin x+ between x nd x π. By solving y sin x + nd y sin x + simultneously, the x-coordinte of their points of intersection re x, x π nd x π. From x to x π, sin x + sin x + while from x π to x π, sin x + sin x +. Hence, the required re: π Exmple 4.. ( sin x + sin x ) + π π (sin x + sin x ) [ cos x] π 4 Find the re of the region bounded by the curve x y nd the line x y [cos x]π π By solving x y nd x y + 6 simultneously, the y-coordinte of their points of intersection re nd. When y, y + 6 y. Hence, the required re: ( y + 6 y ) dy [ y 5 6 ] y + 6y Exmple 4.4. Find the re of the region bounded by the curve y x nd the lines y 5x nd y 4 x. By solving y x nd y 5x simultneously, the x-coordinte of their points of intersection is. Similrly, by solving y x nd y 4 x simultneously, the x-coordinte of their points of intersection is. Hence, the required re: (5x x ) + (4 x x ) [ 5x x ] ] + [4x x x 7 6

33 CHAPTER 4. APPLICATION OF INTEGRATION 4.4 Volume of Solid of Revolution The volume of solid obtined by revolving the region bounded by curve y f(x), the x-xis nd the lines x nd x b bout the xis is given by π b [f(x)]. Similrly, the volume of solid obtined by revolving the region bounded by curve x g(y), the y-xis nd the lines y nd y b is given by Exmple 4.5. π b [f(y)] dy. Find the volume of the solid of revolution generted by revolving the region bounded by the x-xis nd the lines y x +, x nd x 5 bout the xis. The required volume: π 5 [ (x + ) (x + ) π 8π ] 5 Exmple 4.6. Find the volume of the solid of revolution generted by revolving the region bounded by the y-xis nd the lines y x + nd y 7 bout the y-xis. The required volume: π 7 (y ) dy [ (y ) π 7π ] 7 Exmple 4.7. Find the volume of the solid generted by revolving the region bounded by () the curve y x nd the lines y, x nd x bout the line y, (b) the curve y x nd the lines x, y nd y 4 bout the line x. () The required volume: π (x ) π (x 4 x + ) [ x 5 π 5 x + x 496π 5 ]

34 CHAPTER 4. APPLICATION OF INTEGRATION 4 (b) Since x y, the required volume: Exmple 4.8. π 4 (y ) dy π π 7π 6 4 (y y + ) dy [ y 4y + y Find the volume of the solid generted by revolving the region bounded by the curve x y nd the line x, y nd y 4 bout the y-xis. ] 4 The required volume: π 4 [ ( y) ] 4 dy π (y ) dy [ y π y ] 4 9π Exmple 4.9. Find the volume of the solid generted by revolving the region bounded by the curve y x + nd the line y x, x nd x bout the x-xis. The required volume: π [ ( ) ] x + x π ( ) x x + x [ x 5 π 45 x 9 + x 7π 5 ] Exercises. A toy cr initilly t rest moves long stright line from point O. Its velocity v ms fter t seconds is given by v t + 5. Let s m be the displcement of the toy cr from O. () Express s in terms of t. (b) Find the displcement of the toy cr from O fter seconds.

35 CHAPTER 4. APPLICATION OF INTEGRATION 5. There is n pple on tree. The pple is 4 m bove the ground. Polly throws bll verticlly upwrds towrds the pple from the ground. The velocity v ms of the bll fter t seconds is given by v 7 t. Cn the bll hit the pple? Explin your nswer briefly.. The ccelertion ms of moving object t time t s is given by cos t. The velocity v ms of the object is.5 ms when t. () Express v in terms of t. (b) Find the mximum velocity of the object. 4. At ny point (x, y) of curve, eqution of the curve. d y 4( x ). If the slope of the curve t (, 7) is 4, find the 5. At ny point (x, y) of curve, d y 48x. If (, ) nd (, 8) lie on the curve, find the eqution of the curve. 6. At ny point (x, y) of curve, d y kx, where k is constnt. The curve ttins its mximum t (, 5) nd its y-intercept is. () Find the vlue of k. (b) Find the eqution of the curve. (c) Find the minimum point of the curve. (d) Sketch the curve for x. 7. HKCEE 5 Additionl Mthemtics Q () Show tht d [x(x + )n ] (x + ) n [(n + )x + ], where n is rtionl number. (b) The slope of ny point (x, y) of curve C is given by dy (x + )4 (6x + ). If C psses through the point (, ), find the eqution of C. 8. HKCEE 6 Additionl Mthemtics Q The slope t ny point (x, y) of curve is given by dy + cos x. If the curve psses through the ( π point 4, π ), find its eqution Two curves y f(x) nd y g(x) intersect t point A, where f(x) x +x+ nd g(x) f(x 4). () Find the coordintes of A. (b) Find the re of the region bounded by the two curves nd the x-xis.. The curves y x + 4 nd y x x both pss through point A on the x-xis. () Find the coordintes of A. (b) Find the re of the region bounded by the two curves nd y-xis.. The curve y x nd the stright line x y + intersect t point A. () Find the coordintes of A. (b) Find the re of the region bounded by the curve, the stright line nd the y-xis.. The curve y cuts the stright lines y x nd x e t P nd Q respectively. x () Find the coordintes of P nd Q. (b) Find the re of the region bounded by the curve y, the x-xis nd the lines y x nd x e. x

36 CHAPTER 4. APPLICATION OF INTEGRATION 6. Find the re in the first qudrnt bounded by the curves y e x, y e x nd x. 4. Find the re bounded by the curve y x nd the stright line x + y Let f(x) p x x + q for ll x q nd g(x) for ll x 4, where p nd q re positive constnts. x + q 4 x The y-intercept of the curve y f(x) is 5 nd the x-intercept of the curve y g(x) is. () Find the vlues of p nd q. (b) Find the point of intersection of two curves. (c) If the re of the shded region bounded by the two curves, the y-xis nd the line x k is 6 ln, where < k <, find the vlue of k. b 6. () Evlute y dy. (b) Find the re bounded by the lower hlf of the ellipse x + y nd the stright line y (b h) b where < h < b. 7. Find the volume of the solid generted by revolving the region bounded by the curve x 4 y nd the line x y + bout the y-xis. 8. Find the volume of the solid generted by revolving the region bounded by the curves y x nd 8y x bout the x-xis. 9. Show, by using integrtion, tht the volume of sphere with rdius r is 4 πr.. Find the volume of the solid of revolution generted by revolving the region bounded by the x-xis nd the curves y x nd y 6 x bout the x-xis.. Find the volume of the solid of revolution generted by revolving the region bounded by the x-xis, x, x π, y sin x nd y cos x bout the x-xis.. Given two curves y f(x) nd y g(x), where f(x) x 6x + nd g(x) f(x) + 4. () Find the points of intersection of the two curves. (b) Find the re of the shded region bounded by the two curves. (c) Find the volume of the solid of revolution generted by revolving the shded region bout the y-xis.

Definite integral. Mathematics FRDIS MENDELU

Definite integral. Mathematics FRDIS MENDELU Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

More information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

Chapters 4 & 5 Integrals & Applications

Chapters 4 & 5 Integrals & Applications Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

More information

Math 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas

Math 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the continuous

More information

MA 124 January 18, Derivatives are. Integrals are.

MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

F (x) dx = F (x)+c = u + C = du,

F (x) dx = F (x)+c = u + C = du, 35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

1 Techniques of Integration

1 Techniques of Integration November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.

More information

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx... Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

The Riemann Integral

The Riemann Integral Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function

More information

Chapter 8: Methods of Integration

Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

More information

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones. Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5. - 5.3) Remrks on the course. Slide Review: Sec. 5.-5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description

More information

Sections 5.2: The Definite Integral

Sections 5.2: The Definite Integral Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

More information

Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).

Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher). Test 3 Review Jiwen He Test 3 Test 3: Dec. 4-6 in CASA Mteril - Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 14-17 in CASA You Might Be Interested

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

We divide the interval [a, b] into subintervals of equal length x = b a n

We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve. Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F

More information

Main topics for the Second Midterm

Main topics for the Second Midterm Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite

More information

Big idea in Calculus: approximation

Big idea in Calculus: approximation Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

Overview of Calculus I

Overview of Calculus I Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

More information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

More information

MATH1013 Tutorial 12. Indefinite Integrals

MATH1013 Tutorial 12. Indefinite Integrals MATH Tutoril Indefinite Integrls The indefinite integrl f() d is to look for fmily of functions F () + C, where C is n rbitrry constnt, with the sme derivtive f(). Tble of Indefinite Integrls cf() d c

More information

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

More information

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all 3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

5.5 The Substitution Rule

5.5 The Substitution Rule 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

More information

Section Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?

Section Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled? Section 5. - Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles

More information

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics Semester 1, 2002/2003 MA1505 Math I Suggested Solutions to T. 3

NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics Semester 1, 2002/2003 MA1505 Math I Suggested Solutions to T. 3 NATIONAL UNIVERSITY OF SINGAPORE Deprtment of Mthemtics Semester, /3 MA55 Mth I Suggested Solutions to T. 3. Using the substitution method, or otherwise, find the following integrls. Solution. ) b) x sin(x

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information

Math 554 Integration

Math 554 Integration Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

More information

1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x

1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)

More information

Math 116 Calculus II

Math 116 Calculus II Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

More information

Review of basic calculus

Review of basic calculus Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

More information

5.2 Volumes: Disks and Washers

5.2 Volumes: Disks and Washers 4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of cross-section or slice. In this section, we restrict

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

MAT187H1F Lec0101 Burbulla

MAT187H1F Lec0101 Burbulla Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

APPLICATIONS OF THE DEFINITE INTEGRAL

APPLICATIONS OF THE DEFINITE INTEGRAL APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through

More information

MATH 144: Business Calculus Final Review

MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

More information

The practical version

The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

More information

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

More information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

More information

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems

More information

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]

More information

7.2 Riemann Integrable Functions

7.2 Riemann Integrable Functions 7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous

More information

Integrals - Motivation

Integrals - Motivation Integrls - Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but

More information

Final Exam - Review MATH Spring 2017

Final Exam - Review MATH Spring 2017 Finl Exm - Review MATH 5 - Spring 7 Chpter, 3, nd Sections 5.-5.5, 5.7 Finl Exm: Tuesdy 5/9, :3-7:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.

More information

APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

More information

1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.

1.2. Linear Variable Coefficient Equations. y + b ! = a y + b  Remark: The case b = 0 and a non-constant can be solved with the same idea as above. 1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt

More information

Review of Riemann Integral

Review of Riemann Integral 1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

More information

Calculus II: Integrations and Series

Calculus II: Integrations and Series Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

More information

Fundamental Theorem of Calculus

Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

cos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves

cos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. U-Substitution for Definite Integrls A. Th m 6-Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the

More information

10 Vector Integral Calculus

10 Vector Integral Calculus Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

Distance And Velocity

Distance And Velocity Unit #8 - The Integrl Some problems nd solutions selected or dpted from Hughes-Hllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl

More information

Sample Problems for the Final of Math 121, Fall, 2005

Sample Problems for the Final of Math 121, Fall, 2005 Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.

More information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

Calculus I-II Review Sheet

Calculus I-II Review Sheet Clculus I-II Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Topics Covered AP Calculus AB

Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.

More information

Week 10: Riemann integral and its properties

Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

More information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

Chapter 8.2: The Integral

Chapter 8.2: The Integral Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in

More information

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time) HIGHER SCHOOL CERTIFICATE EXAMINATION 999 MATHEMATICS UNIT (ADDITIONAL) Time llowed Three hours (Plus 5 minutes reding time) DIRECTIONS TO CANDIDATES Attempt ALL questions ALL questions re of equl vlue

More information

Unit 5. Integration techniques

Unit 5. Integration techniques 18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble

More information

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus 7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e

More information

38 Riemann sums and existence of the definite integral.

38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

More information

20 MATHEMATICS POLYNOMIALS

20 MATHEMATICS POLYNOMIALS 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

More information