1 Techniques of Integration


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1 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. Suppose tht g(x) is differentible function nd f is continuous on the rnge of g. Integrtion by substitution is given by the following formuls: Indefinite Integrl Version: f(u) du = Definite Integrl Version: g(b) g() f(u) du =. Integrtion By Prts f(g(x))g (x) dx b f(g(x))g (x) dx where u = g(x). where u = g(x). We cn think of integrtion by substitution s the counterprt of the product rule for differentition. Suppose tht u(x) nd v(x) re continuously differentible functions. Integrtion by prts is given by the following formuls: Indefinite Integrl Version: u(x)v (x) dx = u(x)v(x) Definite Integrl Version: b u(x)v (x) dx = u(x)v(x).3 Averge Vlue of Function x=b x= The verge vlue of function f on the intervl [, b] is given by f ve = b b u (x)v(x) dx. b u (x)v(x) dx. We cn interpret f ve s the number such tht hlf the re of the curve lies bove it, nd hlf the re lies below it. In other words, f ve is the number such tht b (f(x) f ve ) dx =. Exmple: The verge of f(x) = x nd g(x) = x + on the intervl [, ] is displyed below: 4 3 Figure : f ve = 4 x dx = 4 3 Figure : g ve = 4 x + dx = Pge of
2 November 8, 8 MAT86 Week Justin Ko.4 More Properties of Integrtion. Integrtion of Odd Functions: If f(x) is odd, then f(x) dx =.. Integrtion of Even Functions: If f(x) is even, then f(x) dx = 3. The following theorem sys tht continuous function ttins its verge vlue: Theorem (The Men Vlue Theorem for Integrls). If f is continuous on [, b], then there exists number c [, b] such tht f(c) = f ve. Tht is, there exists c [, b] such tht b f(x) dx = f(c)(b ). 4. We cn move the verge integrl inside of function tht is concve up: Theorem (Jensen s Inequlity). If f(x) is continuous function on [, b] nd g (x) for ll x in the rnge of f, then g(f ve ) (g f) ve. Tht is,.5 Exmple Problems ( g b.5. Integrtion by Substitution b ) f(x) dx b b g(f(x)) dx. Strtegy: The ide is to mke the integrl esier to compute by doing chnge of vribles.. Strt by guessing wht the pproprite chnge of vrible u = g(x) should be. Usully you choose u to be the function tht is inside the function.. Differentite both sides of u = g(x) to conclude du = g (x)dx. If we hve definite integrl, use the fct tht x = u = g() nd x = b u = g(b) to lso chnge the bounds of integrtion. 3. Rewrite the integrl by replcing ll instnces of x with the new vrible nd compute the integrl or definite integrl. 4. If you computed the indefinite integrl, then mke sure to write your finl nswer bck in terms of the originl vribles. Problem. ( ) Find the following indefinite integrl e x e x tnh(x) dx = e x dx + e x Pge of
3 November 8, 8 MAT86 Week Justin Ko Solution. Step : We will use the chnge of vribles u = e x + e x, du dx = ex e x du = (e x e x ) dx. Step : We cn now evlute the integrl under this chnge of vribles, e x e x du e x dx = = ln u + C + e x u = ln e x + e x + C. u = e x + e x Since e x + e x >, we cn remove the bsolute vlues if we wish giving the finl nswer tnh(x) dx = ln(e x + e x ) + C. Remrk: We cn use the fct e x + e x = cosh(x) to conclude tht ln(e x + e x ) + C = ln( cosh(x)) + C = ln(cosh(x)) + ln() + C = ln(cosh(x)) + D. }{{} D This form of the indefinite integrl my be esier to remember since it mirrors the fct tht tn(x) dx = ln cos(x) + C. Problem. ( ) Evlute the following integrl x x dx. Solution. Step : We will use the chnge of vribles u = x, du dx = x du = x dx du = xdx, x = u =, x = u =. Step : We cn now evlute the integrl under this chnge of vribles, x x dx = u du = 3 u 3 Remrk: Insted of chnging the bounds of integrtion, we cn first find the indefinite integrl, x x dx = ( x ) 3, then use the fundmentl theorem of clculus to conclude x x dx = ( x ) 3 x= x= u= u= = 3. = 3. Pge 3 of
4 November 8, 8 MAT86 Week Justin Ko Problem 3. ( ) Find the following indefinite integrl + x dx. Solution 3. Step : We will use the chnge of vribles u = x. Then du dx = x xdu = dx u du = dx. Step : We cn now evlute the integrl under this chnge of vribles, + x dx = u + u du. This integrl is bit tricky to compute, so we hve to use lgebr to simplify it first. Using long division to first simplify the integrnd, we get u + u du = u + u du = + u du = u ln + u + C = x ln + x + C. u = x. Problem 4. ( ) Find the following indefinite integrl sec(x) dx. Solution 4. We first do trick by multiplying the numertor nd denomintor by sec(x) + tn(x), sec(x)(sec(x) + tn(x)) sec(x) dx = dx = sec(x) + tn(x) Step : We will use the chnge of vribles u = sec(x) + tn(x), sec (x) + sec(x) tn(x) sec(x) + tn(x) du dx = sec(x) tn(x) + sec (x) du = (sec(x) tn(x) + sec (x)) dx. Step : We cn now evlute the integrl under this chnge of vribles, sec (x) + sec(x) tn(x)) sec(x) dx = dx = sec(x) + tn(x) u du = ln u + C dx. = ln sec(x) + tn(x) + C. u = sec(x) + tn(x) Problem 5. ( ) Find the following indefinite integrl sech(x) dx = e x dx. + e x Pge 4 of
5 November 8, 8 MAT86 Week Justin Ko Solution 5. Step : We will use the chnge of vribles u = e x, du dx = ex dx = e x du dx = u du. Step : We cn now evlute the integrl under this chnge of vribles, sech(x) dx = e x dx = + e x u(u + u ) du = u + du = tn (u) + C = tn (e x ) + C. u = e x Alterntive Solution: We first do trick by multiplying the numertor nd denomintor by e x, e x sech(x) dx = e x dx = + e x e x + dx. Step : We will use the chnge of vribles u = e x, du dx = ex du = e x dx. Step : We cn now evlute the integrl under this chnge of vribles, e x sech(x) dx = e x + dx = u + du = tn (u) + C.5. Integrtion by Prts = tn (e x ) + C. u = e x We will introduce method to bookkeep multiple integrtion by prts steps simultneously. This is clled the tbulr method for integrtion by prts. You pick term to differentite nd term to integrte then repet the opertion until product of the terms in the lst entry of the tble is esy to integrte. The integrl cn be recovered by multiplying digonlly cross the rows of the tble dding up ll terms with lternting signs. The lst term in the tble is integrted cross. For exmple, the formul to integrte u(x)v (x) dx by prts cn be encoded by the tble + u v + u v + u v + u v Pge 5 of
6 November 8, 8 MAT86 Week Justin Ko which gives us the formul u(x)v (x) dx = u(x)v (x) u (x)v (x) + u (x)v(x) u (x)v(x) dx..6 Exmples Problem. ( ) Compute xe x dx. Solution. Step : Drw the tble Step : From the tble, we hve + x e x e x + e x xe x dx = xe x e x + C. Problem. ( ) Compute x 6 e x dx. Solution. Step : Drw the tble + x 6 e x 6x 5 e x + 3x 4 e x x 3 e x + 36x e x 7x e x + 7 e x e x Step : From the tble, we hve x 6 e x dx = x 6 e x 6x 5 e x + 3x 4 e x x e x 7xe x + 7e x + C. Pge 6 of
7 November 8, 8 MAT86 Week Justin Ko Problem 3. ( ) Compute x 4 sin x dx. Solution 3. Step : Drw the tble + x 4 sin x 4x 3 cos x + x sin x 4x cos x + 4 sin x cos x Step : From the tble, we hve x 4 sin x dx = x 4 cos x + 4x 3 sin x + x cos x 4x sin x 4 cos x + C. Problem 4. ( ) Compute e x sin x dx. Solution 4. Step : Drw the tble + sin x e x cos x e x + sin x e x Step : From the tble, we hve e x sin x dx = e x sin x e x cos x e x sin x dx + D. Moving ll the e x sin x dx to one side nd simplifying, we cn conclude e x sin x dx = e x sin x e x cos x + D = e x sin x dx = ex sin x ex cos x + C. Problem 5. ( ) Compute xe x cos(x) dx. Pge 7 of
8 November 8, 8 MAT86 Week Justin Ko Solution 5. Step : Drw the tble + x cos x e x cos x x sin x e x + sin x x cos x e x Step : From the tble, we hve xe x cos x dx = xe x cos x e x cos x + xe x sin x e x sin x dx xe x cos x dx. Moving ll the xe x cos x dx to one side nd simplifying, we cn conclude xe x cos x dx = xe x cos x e x cos x + xe x sin x e x sin x dx = xe x cos x e x cos x + xe x sin x e x sin x + e x cos x + C. Problem 4 Dividing both sides by, we cn conclude xe x cos x dx = ( ) xe x cos x + xe x sin x e x sin x + C. Problem 6. ( ) Compute ln(x) dx. Solution 6. Step : Drw the tble + ln(x) x Step : From the tble, we hve ln(x) dx = x ln(x) x dx = x ln(x) x + C. Problem 7. ( ) Compute x 3 ln x dx. Solution 7. Step : Drw the tble Pge 8 of
9 November 8, 8 MAT86 Week Justin Ko + ln x x 3 x Step : From the tble, we hve x 3 ln x dx = 4 x4 ln x 4 4 x4 x 3 dx = 4 x4 ln x 6 x4 + C. Step 3: We cn now use the fundmentl theorem of clculus to compute the definite integrl, x 3 ln x dx = 4 x4 ln x 6 x4 x= x= = 4 ln + 5 = 4 ln 6 6. Problem 8. ( ) Prove the reduction formul sin n (x) dx = n sinn (x) cos(x) + n n sin n (x) dx. Solution 8. Step : Drw the tble + sin n (x) sin(x) (n ) cos(x) sin n (x) cos(x) Step : From the tble, we hve sin n (x) dx = sin n (x) cos(x) + (n ) = sin n (x) cos(x) + (n ) = sin n (x) cos(x) + (n ) cos (x) sin n (x) ( sin (x)) sin n (x) sin (x) + cos (x) = sin n (x) dx (n ) sin n (x) dx Moving ll the the sin n (x) dx terms to one side, we cn conclude n sin n (x) dx = sin n (x) cos(x) + (n ) sin n (x) dx = n sinn (x) cos(x) + n n.6. Proofs of Properties of Integrtion Problem. ( ) Suppose tht f( x) = f(x). Prove tht f(x) dx = sin n (x) dx sin n (x) dx. Pge 9 of
10 November 8, 8 MAT86 Week Justin Ko Solution. By the properties of definite integrls, we hve f(x) dx = Using the chnge of vribles u = x on the first integrl, for even function f, = = This computtion implies f( u) du f(u) du u = x, du = dx, x = u =, x = u = f( x) = f(x) f(x) dx = f(x) dx = Problem. ( ) Suppose tht f( x) = f(x). Prove tht f(x) dx =. Solution. By the properties of definite integrls, we hve f(x) dx = Using the chnge of vribles u = x on the first integrl, for odd functions f, = = This computtion implies f( u) du f(u) du u = x, du = dx, x = u =, x = u = f( x) = f(x) f(x) dx =. Problem 3. ( ) Justify the technique used to solve seprble ordinry differentil equtions: dy dy = f(x)g(y) = dx g(y) = f(x) dx G(y) = F (x) + C where G(y) is n ntiderivtive of g(y) nd F (x) is n ntiderivtive of f(x). Pge of
11 November 8, 8 MAT86 Week Justin Ko Solution 3. Using the nottion y (x) = dy dx dy dx = f(x)g(y) nd writing y = y(x) explicitly s function of x, we hve y (x) g(y(x)) dx = y (x) g(y(x)) = f(x) Using the chnge of vribles u = y(x) on the first integrl involving the y(x) term, we see y (x) g(y(x)) dx = du g(u) = dy g(y). Therefore, using this chnge of vribles, we cn conclude tht dy dy = f(x)g(y) = dx g(y) = This mens there is hidden chnge of vribles tht goes on when we formlly seprted dy dx in the second step of the technique. Problem 4. ( ) Vnishing Theorem: Suppose tht f(x) is continuous function on [, b] such tht f(x) for ll [, b] nd Prove tht f(x) = for ll x [, b]. b f(x) dx =. Solution 4. On the contrry, suppose tht f(x ) > for some point x [, b]. Then by continuity, we must lso hve tht f(x) > on some intervl [k, l] [, b]. By the men vlue theorem of integrtion, there exists c [k, l] such tht l k f(x) dx = f(c)(l k). Since we lso hve tht f(x) > for ll x [k, l], we must hve f(c) >, which implies tht l k f(x) dx = f(c)(l k) >. Since f(x), by the monotonicity of integrtion, the conclusion bove implies b f(x) dx l k f(x) dx >, which contrdicts the fct tht b f(x) dx =. Therefore, we must hve tht f(x) = for ll x [, b]. Problem 5. ( ) Prove the Men Vlue Theorem for Integrtion by pplying the Men Vlue Theorem to the function F (x) = x f(t) dt. Solution 5. By the fundmentl theorem of clculus, F (x) is continuous on [, b] nd F (x) is differentible on (, b). Therefore, by the men vlue theorem there exists c (, b) such tht F (b) F () b Since F (b) F () = b f(x) dx f(x) dx = b b f(x) dx b = f(c) = = F (c) = f(c). f(x) dx, there exists c (, b) such tht b Pge of f(x) dx = f(c)(b ).
12 November 8, 8 MAT86 Week Justin Ko Problem 6. ( ) Jensen s Inequlity: Suppose tht f(x) is continuous function on [, b] nd g (x) for ll x. Prove tht ( g b b ) f(x) dx b b g(f(x)) dx. Solution 6. Let L(x) be the liner pproximtion of g t the point x = f ve. Since g (x), we hve g(x) lies bove its tngent line so, L(x) := g(f ve ) + g (f ve )(x f ve ) g(x). Since this holds for ll x, we hve L(f(x)) g(f(x)). Tking the verge integrl of both sides nd using the monotonicity of definite integrls, we cn conclude g(f ve ) = b b g(f ve ) + g (f ve )(f(x) f ve ) dx b b g(f(x)) dx. Remrk: We only require g (x) on the rnge of f for the inequlity to hold. This is becuse f ve is in the rnge of f by the men vlue theorem of integrtion, nd g is only evluted t points in the rnge of f. If g (x) >, then the sme proof shows g(f ve ) < (g f) ve provided tht f is not constnt function. Remrk: It is esy to see tht the opposite inequlity holds if g (x). Suppose tht g (x), then g (x), so Jensen s inequlity implies tht g(f ve ) ( g f) ve = g(f ve ) (g f) ve. Pge of
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