Indefinite Integral. Chapter Integration  reverse of differentiation


 Dustin Nelson
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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 rightngled 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 xxis 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 yxis 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 xxis 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 xxis nd the lines x nd x. x Notice tht the region is below the xxis 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 xxis. When y, x, or. Hence, the xintercepts of the curve re, nd. Notice tht from x to x, the re region is bove the xxis while from x to x, the re region is below the xxis. 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 xxis bounded by the curve x + y 6 nd the line x. Notice tht y ± 6 x nd the required re is below the xxis. 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 yxis. Notice tht we re now considering the re with respect to the yxis insted of the xxis. When x, y, or. Hence, the yintercepts of the curve re, nd. Moreover, the required re is leftwrd of the yxis from y to y nd rightwrd of the yxis 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 xcoordinte 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 xcoordinte 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 ycoordinte 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 xcoordinte of their points of intersection is. Similrly, by solving y x nd y 4 x simultneously, the xcoordinte 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 xxis 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 yxis 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 xxis 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 yxis nd the lines y x + nd y 7 bout the yxis. 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 yxis. ] 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 xxis. 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 yintercept 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 xxis.. The curves y x + 4 nd y x x both pss through point A on the xxis. () Find the coordintes of A. (b) Find the re of the region bounded by the two curves nd yxis.. 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 yxis.. 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 xxis 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 yintercept of the curve y f(x) is 5 nd the xintercept 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 yxis 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 yxis. 8. Find the volume of the solid generted by revolving the region bounded by the curves y x nd 8y x bout the xxis. 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 xxis nd the curves y x nd y 6 x bout the xxis.. Find the volume of the solid of revolution generted by revolving the region bounded by the xxis, x, x π, y sin x nd y cos x bout the xxis.. 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 yxis.
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