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 9 Compute 6 x 5 dx by finding s of regions between the grph of f nd the x-xis. Solution to Problem 9 x-intercept 0 = x 5 x = 5 ( under f bove x-xis) A = (6 5 ) ( 6 5) = 49 4
orem of definite Solution to Problem 9 (continued) ( bove f under x-xis) 6 A = ( 5 ( )) ( ( ) + 5) = ( 9 ) (9) = 8 4 x 5 dx = A A = 49 4 8 4 = 8 Exmple 0 (Are formuls seldom work) Evlute 3 5 5 x dx. No formul for this portion of circle
orem of definite orem ( orem of I) If f is continuous on the intervl [, b] nd F (t) = f (t) then b f (t) dt = F (b) F (). Exmple ( ) sin 8x + x 4 = 8 cos 8x + 4x 3 so by FTOC d dx 8 cos 8x + 4x 3 dx = (sin(8 ) + 4 ) (sin(8 ) + 4 ) = sin 6 sin 8 + 5
orem of definite Nottion g(x) b = g(b) g() x= or if x is clerly the vrible to plug nd b into cn write g(x) b = g(b) g() With this nottion FTOC I ( orem of I) sys If f is continuous on the intervl [, b] nd F (t) = f (t) then b f (t) dt = F (t) b t=
orem of definite FTOC sys roughly Computing under f Suprising! Why should nd slopes be relted? Finding with derivtive f Could prove FTOC from definition of integrl (Definition 9) nd definition of the derivtive, but we won t. Mkes some sense visully
orem of definite Note on units for If x mesured in units u nd f (x) mesured in units u then hs units u u Exmple 3 b f (x) dx If t is the time mesured in hours P(t) number of people working in fctory t time t. b P(t) dt is people hours worked between time nd b
orem of definite Integrl of rte of chnge If F (t) is the rte of chnge of some quntity F (t) then FTOC I sy tht b F (t) dt = F (b) F () This is net chnge in F from time to b Exmple 4 As in Exmple 4 from lst week t time (in h) v(t) velocity (rte of chnge of position) t time t (in km/h) then 3 0 v(t) dt is distnce trveled (net chnge in position) in units (km/h) h = km
orem of definite Exmple 5 t time (in yers) g(t) growth rte (in m/yer) of person t ge t then 6 9 g(t) dt is chnge in height (in m) of person from ge 9 to 6.
orem of definite Definition 6 ( ) verge vlue of f on the intervl [, b] is f ve = b Where does this formul come from? Let s derive it using b f (t) dt. Philosophy of Estimte quntity Figure out how to improve estimte Tke limit
orem of definite Averge temperture Let f (t) be the temperture t time t. Let s estimte the verge temperture between m nd 0pm. (First estimte with intervls) totl time = 0 ech intervl is 0 f ve f ( + 0 (Better estimte with 3 intervls) totl time = 0 ech intervl is 0 3 f ve f ( + 0 3 ) ( + f + 0 3 3 ) ( ) + f + 0 ) ( ) + f + 3 0 3
orem of definite Averge temperture (continued) (Better estimte with n intervls) totl time = 0 ech intervl is 0 n f ve f ( + 0 n = n f i= = 0 ( + i 0 n n f i= ) ( ) + + f + n 0 n n ) n ( + i 0 n ) 0 n
orem of definite Averge temperture (continued) (Tke limit to get exct verge) f ve = lim n 0 = 0 lim n = 0 0 n f i= n f i= f (t) dt ( + i 0 n ( + i 0 n ) 0 n ) 0 n We ve derived the formul in Definition 6.
orem of definite Problem 7 Find the verge vlue of the funtion on the intervl [, ] Solution to Problem 7 h ve = b = h(x) = 4 x b ( ) = 4 π h(x) dx 4 x dx = π
orem of definite Problem 8 t is time mesured in dys since Jn., 003 R(t) is the distnce from the erth to the sun t time t Wht does represent? Solution to Problem 8 365 0 365 0 R(t) dt Averge distnce from erth to the sun in 003
orem of definite orem 9 If f is integrble then orem 30 If f is integrble then definite b f (x) dx = b f (x) dx b c c f (x) dx + f (x) dx = f (x) dx b
orem of definite orem 3 If f is integrble then Proof. b b cf (x) dx = c cf (x) dx = lim n = lim n c b f (x) dx n cf ( + i x) x i= = c lim n = c b n f ( + i x) x i= n f ( + i x) x i= f (x) dx
orem of definite orem 3 If f nd g re integrble then Proof. b b f (x) + g(x) dx = f (x) + g(x) dx = lim n = lim n b f (x) dx + b g(x) dx n (f ( + i x) + g( + i x)) x i= n (f ( + i x) x + g( + i x) x) i=
orem of definite Proof of orem 3 (continued). n = lim (f ( + i x) x + g( + i x) x) n i= ( n ) n = lim f ( + i x) x + g( + i x) x n i= i= ( ) ( ) n n = f ( + i x) x + g( + i x) x = lim n b i= f (x) dx + b g(x) dx lim n i=
orem of definite Problem 33 Compute 6 7 t(v) 7u(v) dv given 6 7 u(v) dv = 6 7 t(v) dv = Solution to Problem 33 6 7 t(v) 7u(v) dv = = 6 7 6 7 t(v) dv t(v) dv 7 = 7( ) = 6 6 7 6 7 7u(v) dv u(v) dv
Wrning orem of definite Exmple 34 f (x) = b f (x) g(x) dx b {, 0 x 0, < x b f (x) g(x) dx f (x) dx b b f (x) dx g(x) dx nd g(x) = b g(x) dx { 0, 0 x, < x f (x)g(x) dx = 0 0 0 dx = 0 but f (x) dx 0 0 g(x) dx = =
orem of definite Definition 35 f is even if f ( x) = f (x) f is odd if f ( x) = f (x) Note: Most functions re neither. One function is both. orem 36 If f is even then If f is odd then nd 0 f (x) dx = 0 0 f (x) dx = 0 f (x) dx = f (x) dx f (x) dx = 0 f (x) dx
orem of definite Problem 37 Suppose 4 Q(s) ds = 3 0 Q(s) ds = 9 4 Evlute 0 Solution to Problem 37 0 Q(s) ds = 4 Q(s) ds Q(s) ds + = 3 + ( 9) = 6 0 4 Q(s) ds
orem of definite Problem 38 Compute 8 y(u) du given y is n odd function 8 y(u) du = 7 Solution to Problem 38 8 8 y(u) du = y(u) du y(u) du = 7 0 = 7
orem of definite Problem 39 Compute 7 h(r) dr given 3 h is n even function 7 h(r) dr = 4 0 7 h(r) dr = 0 3 Solution to Problem 39 7 3 h(r) dr = = = 7 3 7 3 7 = 3 7 h(r) dr h(r) dr 3 3 0 3 ( 7 h(r) dr 3 h(r) dr + h(r) dr h(r) dr h(r) dr 3 7 = ( 0) + (4) = 8 0 h(r) dr 7 0 ) h(r) dr
orem of definite orem 40 If m f (x) M for x [, b] then m(b ) orem 4 b If f (x) g(x) for x [, b] then f (x) dx M(b ) b f (x) dx b g(x) dx
orem of definite Problem 4 Show tht 99 00 cos(x) sin(x 3 ) dx 99 Solution to Problem 4 For ll x [, 00] cos(x) sin(x 3 ) = cos(x) sin(x 3 ) so cos(x) sin(x 3 ) By orem 40 (00 ) 99 00 00 cos(x) sin(x 3 ) dx (00 ) cos(x) sin(x 3 ) dx 99
orem of definite Problem 43 For ech pirs of decide which is the lrger π 4 0 cos(x) dx nd π 4 sin(x) dx 0 π π 4 cos(x) dx nd π π 4 Solution to Problem 43 sin(x) dx For ll x [0, π 4 ] cos(x) sin(x) so π 4 0 cos(x) dx π 4 0 sin(x) dx For ll x [ π 4, π ] cos(x) sin(x) so π π 4 cos(x) dx π π 4 sin(x) dx
orem of definite Definition 44 (An ntiderivtive) F (x) is n ntiderivtive of f (x) if F (x) = f (x). Exmple 45 so ( d dx x cos(4x + 3) ) = x cos(4x + 3) + 4x sin(4x + 3) x cos(4x + 3) is n ntiderivtive of x cos(4x + 3) + 4x sin(4x + 3) x cos(4x + 3) + 30 is nother ntiderivtive.
orem of definite Motivtion Why do we cre bout finding? FTOC I sys tht computing b f (t) dt is esy if we hve n ntiderivtive F of f. Exmple 46 From Exmple 45 bove 7 x cos(4x + 3) + 4x sin(4x + 3) dx = x cos(4x + 3) + 30 7 = (7 cos(4 7 + 3) + 30) ( cos(4 + 3) + 30) = 49 cos(3) cos(7)
orem of definite Notice in Exmple 45 we could hve dded ny constnt to x cos(4x + 3) nd we would hve hd nother ntiderivtive of x cos(4x + 3) + 4x sin(4x + 3) We usully dd n unspecified constnt to remind us tht there re mny. Definition 47 ( ntiderivtive) ntiderivtive of f (x) is the set of ll of f (x). orem 48 If f is continuous nd F (x) = f (x) then every ntiderivtive of f is of the form F (x) + C for some constnt C.
orem of definite Wht if f is not continuous? ntiderivtive of noncontinuous function Let n F (x) = { ln x + 4, x > 0 ln( x) + 8, x < 0 { F (x) = x, x > 0 { x, x < 0 = x, x > 0 x, x < 0 = x So F (x) is n ntiderivtive of x. Any choice of constnts (4 nd 8 weren t specil) gives sme result. Thus the ntiderivtive of x is F (x) = { ln x + C, x > 0 ln( x) + C, x < 0 = { ln x + C, x > 0 ln x + C, x < 0
orem of definite On the other hnd Min reson we cre bout is the FTOC. FTOC only pplies if f is integrble on [, b] x is not integrble on intervls contining 0 so in pplictions we only use one of the two constnts t time Exmple 49 3 x dx = ln x +C Exmple 50 7 4 x dx 3 = (ln +C ) (ln 3 +C ) = ln ln 3 cnnot be evluted using FTOC Nottionl wrning By convention we sy tht F (x) + C is the ntiderivtive of f (x) whenever F (x) = f (x) even when this is techniclly incorrect.
orem of definite Nottion f (x) dx = F (x) + C mens F (x) + C is the ntiderivtive of f (x) Terminology Since FTOC links ntidifferentition nd integrtion we lso cll (indefinite). following sttements ll men the sme thing: f (x) = d dx F (x) f (x) dx = F (x) + C f (x) is the derivtive of F (x) F (x) + C is the ntiderivtive of f (x) F (x) + C is the indefinite integrl of f (x) F (x) + C is the integrl of f (x)
orem of definite Problem 5 Check the following (6x + 3e x ) cos(3x + 3e x ) dx = sin(3x + 3e x ) + C sec x dx = ln sec x + tn x + C Solution to Problem 5 d dx sin(3x + 3e x ) = (6x + 3e x ) cos(3x + 3e x ) d dx ln sec x + tn x = sec x tn x + sec x sec x + tn x tn x + sec x = (sec x) sec x + tn x = sec x Note: Similr to x ntiderivtive of sec x should hve different C for ech intervl [ (n )π, (n+)π ] but nobody does this.
orem of definite (See bckbord)
orem of definite Ech rule for differentition gives us rule for integrtion From ( ) c d dx F (x) = d dx cf (x) we get orem 5 (Constnt rule for integrtion) cf (x) dx = c f (x) dx
orem of definite Proof of orem 5. Suppose d dx F (x) = f (x). We hve the derivtive rule c d dx F (x) = ( ) d dx cf (x) Reinterpreting this rule s n ntiderivtive gives c d dx F (x) dx = cf (x) + C. Thus we my conclude cf (x) dx = c d dx F (x) dx = cf (x) + C = c(f (x) + C ) = c f (x) dx.
orem of definite From we get ( ) d dx F (x) + d dx G(x) = d dx F (x) + G(x) orem 53 (Sum rule for integrtion) f (x) + g(x) dx = f (x) dx + g(x) dx
orem of definite Proof of orem 53. Suppose d d dx F (x) = f (x) nd dx G(x) = g(x). We hve the derivtive rule ( ) d dx F (x) + d dx G(x) = d dx F (x) + G(x) Reinterpreting this rule s n ntiderivtive gives d dx F (x) + d dx G(x) dx = F (x) + G(x) + C. Thus we my conclude f (x) + g(x) dx = d dx F (x) + d dx G(x) dx = F (x) + G(x) + C = f (x) dx + g(x) dx Note: We drop constnts when we hve on both sides of n eqution.
orem of definite Bsic Integrls Ech bsic derivtive gives us bsic integrl differentition rule Tble : Bsic to memorize integrtion rule d dx x r+ = (r + )x r x r dx = r+ x r+ + C if r d dx ln x = x d dx cos x = sin x d dx sin x = cos x d dx ex = e x d dx rctn x = +x d dx rcsin x = x x dx = ln x + C sin x dx = cos x + C cos x dx = sin x + C e x dx = e x + C +x dx = rctn x + C x dx = rcsin x + C
orem of definite More bsic You lso know few more derivtive Tble : More bsic to memorize differentition rule d dx tn x = sec x d dx cot x = csc x d dx sec x = sec x tn x d dx csc x = csc x cot x d dx x = (ln ) x integrtion rule sec x dx = tn x + C csc x dx = cot x + C sec x tn x dx = sec x + C csc x cot x dx = csc x + C x dx = x ln + C
orem of definite Techniques of integrtion Advnced derivtive give us techniques of integrtion differentition technique of rule integrtion chin rule u-substitution ( 7.) product rule integrtion by prts ( 7.) We will return to these integrtion techniques lter.
orem of definite Problem 54 Find formul for 0e x + 7 sin x dx Solution to Problem 54 0e x + 7 sin x dx = Check your nswer! = 0 0e x dx + e x dx + 7 7 sin x dx (Sum rule) sin x dx (Constnt rule) = 0e x 7 cos x + C (Tble ) d dx (0ex 7 cos x) = 0e x + 7 sin x
orem of definite Problem 55 Find formul for 8 t 8t dt Solution to Problem 55 8 t 8t dt = 8 = 8 dt t dt 8 t 8 t dt (Sum rule) t dt (Const. rule) = 8 rcsin t 8 t3 3 + C (Tble ) Check your nswer! ( d dt 8 rcsin t 8 3 t3) = 8 t 8 3 3t = 8 t 8t
orem of definite Problem 56 Find formul for rcsin(3 π ) Solution to Problem 56 rcsin(3 π ) = + cos u du + sec u du rcsin(3 π ) du + sec u du (Sum rule) = rcsin(3 π ) u + tn u + C (Tbles nd ) Check your nswer! ( d rcsin(3 π ) dt ) u + tn u = rcsin(3 π ) + sec u
orem of definite Problem 57 Compute 3 y( 6 6 y) 3 y dy Solution to Problem 57 3 y( 6 6 y) dy = y 3 = = = 3y ( y 6 + 36y 6 ) dy y 3 3y 36y 4 6 + 08y 5 6 ) dy y 3 3y 6 36y 6 + 08y 3 6 dy 3y 6 36y 3 + 08y dy = 3 6 7 y 7 6 36 3 4 y 4 3 + 08 = 8 7 y 7 4 3 6 7y 3 + 7y = 3 y 3 ( 8 7 7 6 7 4 3 + 7 3 ) ( 8 7 + 45)
orem of definite Problem 58 Find n ntiderivtive G(x) of g(x) = sin x + 7 stisfying G(π) = 0. Solution to Problem 58 G(x) = sin x + 7 dx = cos x + 7x + C Use fct tht G(π) = 0 to solve for C. 0 = G(π) = cos π + C So C = 0 + cos π = 0 + ( ) = G(x) = cos x + 7x
orem of definite Problem 59 verge vlue of h(x) = x 3 3x on [, ] is 8 solve for. Solution to Problem 59 h ve = x 3 3x dx ( ) = ( 4 x 4 x 3 ) = ( ) [ 4 ( )4 ( ) 3 ] ( 4 4 3 ) = ( 4 4 + 3 4 4 + 3) = ( 3) = Use fct tht h ve = 8 to solve for. 8 = h ve = so = ± 8