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


 Clyde Banks
 1 years ago
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1 Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find ntiderivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible function whose rnge is n intervl I nd f is continuous on I, then f(g(x))g (x) dx = f(u) du This method is sometimes lso referred to s usubstitution. This is relly the chinrule for differentition being pplied in reverse order. Exmple: Evlute e 4x dx We note tht the derivtive of e 4x is e 4x, but differentiting e 4x will lso give rise to n dditionl fctor of 4 becuse of the chin rule, so to see this problem better we mke this substitution: u = 4x From this substitution we then hve: du dx = 4 The substitution method sys tht we cn tret du, dx s differentils, in other words, we cn write the bove s: du = 4 dx You cn look t the substitution method s wy of recognizing the ntiderivtive of the result of the chin rule. Treting du, dx s differentils llow us to rewrite every term inside the integrl sign, including dx, to be in terms of u nd du. u = 4x du = 4 dx dx = 4 du ( ) e 4x dx = e u du = e u du = eu + C Since the originl question is in terms of x, our nswer must lso be in terms of x, so we bck substitute to get: e 4x dx = 4 eu + C = 4 e4x + C Notice tht with the substitution method (or simply by inspection), we see tht if the ntiderivtive of function f(x) cn be found, then the ntiderivtive of function of the form f(cx), where c is ny constnt, cn lso be found. In fct, if g(x) is the ntiderivtive of f(x), then the ntiderivtive of f(cx) will be c g(cx) For exmple, since e x dx = e x + C, so e 5x dx = 5 e5x + C
2 Similrly, sin x dx = cos x + C, so sin(8x) dx = 8 cos(8x) + C Keep in mind tht this works only if c is constnt. It is importnt to note tht when you use substitution, you must ccount for everything in the originl vrible. If you substitute u in terms of x, you must lso ccount for dx in terms of u nd du. Most of the time, dx does not just chnge into du, but insted, you will need to differentite nd tret them s differentils/numbers. Exmple: Find x sin(x ) dx Let u = x, then du = x dx nd the bove integrl becomes: x sin(x ) dx = sin(u) du = cos u + c = cos(x ) + c Exmple: x Find x dx Let u = x, then du = x dx, so x dx = du x x dx = u du = u du = ln u + c = ln x + c Exmple: Find tn(x) dx Ans: tn(x) dx = sin(x) cos(x) dx Let u = cos(x), then du = sin(x)dx, so du = sin(x)dx, nd we hve sin(x) cos(x) dx = u du = du = ln u + c = ln cos(x) + c u Using the property of log, we know tht r ln(x) = ln(x r ), the bove expression cn be written s: ( ) ln cos x + C = ln cos x + C = ln + C = ln sec x + C cos x Notice in this exmple tht, when use use the substitution u = cos(x), dx does
3 not just become du. Insted, dx = du. We hve the expression sin x dx sin x inside the integrl nd tht is the reson we chnge everything in the expression to be in terms of u nd du. Exmple: x + x + dx We mke the substition u = x+, then x = u x = (u ), lso, du = dx, so the bove expression becomes: x + (u ) x + dx = + u 4u u 4u + 5 du = du = du u u u We cn now split the frction into three frctions nd simplify: u 4u + 5 u du = u u 4u u + 5 u du ( ) = u du = u 4u + 5 ln u + C u Rewrite the expression bck in terms of x, we get: x + u (x + ) dx = 4u + 5 ln u + C = 4(x + ) + 5 ln x + + C x + When you use substition for definite integrl, you hve two options. You cn either finish the integrtion nd bck substitute the vlues of x using the limits of integrtion, or you cn chnge the limits of integrtion t the beginning. Exmple: e ln x x dx Approch : Let u = ln x, then du = x dx, so e ln x x=e x dx = u du = u x= ] x=e x= = (ln x) ] e = (ln e) (ln ) Notice tht in this pproch, we do not chnge the limits of integrtion when we substitute u for x, so we explicitly remind ouself tht the limits of integrtion re for x, not for u, nd fter we finish the integrtion, we substitute bck in x. Approch : Let u = ln x, then du = dx. Also, since u = ln x, when x =, x u = ln = 0, nd when x = e, u = ln e =, we chnge the limits of integrtion in terms of u: =
4 e ] ln x x dx = u du = u = 0 0 In this pproch the limits of integrtion is immeditely chnged together with the chnge from x to u, nd we do not need to chnge u bck to x once we finish the problem. Either pproch is perfectly vlid nd you my use whichever you like. Integrtion by Prts: Suppose f, g re differentible functions of x, the product rule for differentition gives: (f(x) g(x)) = f (x) g(x) + g (x) f(x) Tking ntiderivtive of both sides nd using the property of integrls, we hve: f(x) g(x) = f (x) g(x) dx + g (x) f(x) dx Rerrnge the terms we get: g (x) f(x) dx = f(x) g(x) g(x) f (x) dx Using u to represent f(x) nd v to represent g(x), the bove formul cn be written in this more compct form: u dv = uv v du The bove formul is the method of integrtion by prts. You cn think of this method s the reverse of the product rule for differentition. Wht this method llows us to do is to chnge the formt of n integrl to nother one which we (hopefully) cn more esily find its derivtive. When pplying the method of integrtion by prts, the originl integrl strts s u dv, where u is function of x, nd dv is nother function of x, including the differentil dx. You then turn u into du, nd find the function v s function of x, given tht you know wht is dv. The integrtion by prts formul then sys you hve to evlute different integrl, v du. Exmple: Find xe x dx If we let u = x nd dv = e x dx, then du = dx nd v = e x, nd the bove integrl becomes:
5 xe x dx = u dv, using the method of integrtion by prts, we get: xe x dx = u dv = uv v du = xe x e x dx = xe x e x + C Exmple: Find ln x dx We wnt to use integrtion by prts to solve this integrl. While the integrl ppers to hve only one product, ln x, don t forget tht for ny expression, dv = dv. We will let u = ln x nd let dv = dx, then du = dx, nd v = x, x nd using integrtion by prts we hve: ln x dx = u dv = uv v du = xln x x x dx = x ln x dx = x ln x x + C Exmple: Evlute x 5 x 3 + dx Let s try to solve this problem by the method of integrtion by prts. First we will decide which fctor inside the integrnd we will use to be u nd which will be dv. If we let u = x 5 nd dv = x 3 + dx, we will hve hrd time figuring out wht is v, since to find dv we will need to first find the ntiderivtive of x 3 +, which is not esy to do. We my try to let u = x 3 + nd dv = x 5 dx. This wy t lest we cn find v
6 using the power rule, nd this gives us: 3x du = x6 dx v = x 3 + 6, Using the formul for by prts we get: x 5 x 3 + dx = x 3 + x6 6 x3 + x 6 6 3x x 3 + dx = x6 x x 3 + dx We end up with n integrl tht is more complicted thn wht we strted with. The trick for this problem is, insted of using ll 5 powers of x s fctor of u, we notice tht the derivtive of x 3 will give rise to fctor of x term (with constnt fctor, but tht is not importnt). Wht we do insted is to redistribute the powers of x by looking t the integrl s: x 5 x 3 + dx = x (x 3 ) x 3 + dx And we let: u = x 3, dv = x x 3 + dx We cn integrte x x 3 + dx by substitution (use z = x 3 + ) to find v, then we hve: du = 3x dx, v = ( x 3 + ) 3/, 9 then the integrl becomes: x 5 x 3 + dx = x (x 3 ) x 3 + dx = u dv = uv v du [ ( = x 3 x 3 + ) ] [ 3/ ( x 3 + ) ] 3/ 3x dx 9 9 The integrl on the right cn now be evluted by substitution gin, nd we get:
7 x 5 x 3 + dx = x (x 3 ) x 3 + dx = u dv = uv v du [ ( = x 3 x 3 + ) ] [ 3/ ( x 3 + ) ] 3/ 3x dx 9 9 = ( 9 x3 x 3 + ) 3/ 4 ( x 3 + ) 5/ + C 45 Exmple: Evlute e x sin x dx We will evlute by prts. In this exmple it does not mtter which one we use s our u nd which one is dv. Let u = e x, then du = e x dx, dv = sin x dx v = cos x from the formul for integrtion by prts we hve: e x sin x dx = u dv = uv v du = e x ( cos x) ( cos x)e x dx = e x cos x + e x cos x dx The integrl on the right seems s complicted s the one we strted with, so wht s the difference here? To see wht hppens, we will use the by prts method gin, still letting u = e x nd this time, dv = cos x dx, then: du = e x dx, v = sin x Using by prts gin we hve: e x sin x dx = e x cos x + e x cos x dx [ ] = e x cos x + e x sin x (sin x)e x dx = e x cos x + e x sin x e x sin x dx
8 We end up with the sme integrl we strted with. It my pper tht we hve gone no where, but notice tht the integrl on the left hs negtive sign, so rerrnging the eqution gives us: e x sin x dx = e x cos x + e x sin x e x sin x dx = e x cos x + e x sin x e x sin x dx = ( ex cos x + e x sin x) + C e x sin x dx Notice tht we still need to put in the +C t the end. Integrls involving Trigonometric Functions: In finding ntiderivtives involving trifunctions, sometimes we need to chnge the expression into different forms using trigonometric identities, nd fter tht, if necessry, use substitution or other methods to finish the integrl. Exmple: Evlute sin 5 x dx This is n integrl tht involves n odd power of sin, we use the fct tht the derivtives of sin x is cos x nd the pythgoren identity: sin x + cos x = to chnge the bove into: sin 5 x dx = (sin 4 x ) (sin x) dx = (sin x ) (sin x) dx = ( cos x ) (sin x) dx Using the substitution u = cos x, we get: du = sin x dx, nd the bove integrl becomes: ( cos x ) (sin x) dx = ( u ) ( du) = ( u + u 4 ) du = + u u 4 du This is now polynomil in u which cn be esily evluted using the power rule: ( ) u + u u 4 3 du = u + u C We now just substitute the expression bck to in terms of x: ( ) u 3 u + u C = cos x + 3 cos3 x cos5 x + C 5 Exmple: sin 6 x dx This time we hve n even power of sine. The previous pproch we used will
9 not work, s when we chnge fctors of sin x to cos x, nd use one fctor sin x to ccount for du, there will be one fctor of sin x left, which will not be ccounted for. Insted, we use nother identity from trigonometry to reduce the powers of sine. Remember the double ngle identity for cosine: cos(x) = cos x = sin x If we use the first eqution nd solve for cos x we get: cos(x) = cos x cos x = cos(x) + cos x = (cos(x) + ) Using the second eqution nd solve for sin x we get: cos(x) = sin x sin x = cos(x) sin x = ( cos(x)) These two formuls re the power reduction or hlf ngle formul from trigonometry. We cn use these formul to reduce power of sine or cosine. With this, the bove gives: (sin sin 6 x dx = x ) ( ) 3 3 dx = ( cos x) dx = 3 cos(x) + 3 cos (x) cos 3 (x) dx 8 = [ dx 3 cos(x) dx + 3 cos (x) dx 8 = [x 3 ( ) 8 sin(x) + 3 ( + cos(4x)) dx ] cos 3 (x) dx ] cos (x) cos(x) dx Notice tht the hlfngle identity is used gin for the third integrl. = [x 3 8 sin(x) + 3 x + 38 ] sin(4x) ( sin (x)) cos(x) dx = [x 3 8 sin(x) + 3 x + 38 ] sin(4x) cos(x) cos(x) sin (x) dx We use the substitution method gin for the lst integrl, with u = sin(x) = [ x 3 8 sin(x) + 3 x + 3 ( 8 sin(4x) sin(x) )] 6 sin3 (x) + c
10 = 5 6 x 4 sin(x) sin(4x) + 48 sin3 (x) + C Applying similr methods, by using the pythgoren identity or the hlfngle identity, we cn integrte ny function of the form sin n x, cos n x, where n could be n odd or even number. Wht if we hve products of sine nd cosine? Exmple: Evlute: sin 3 x cos 6 x dx If we hve fctors of both sine nd cosine in our integrl, s long s one of the powers is odd, we cn tke one of the fctors to be the derivtive of the other function, nd mke usubstitution with u to be the other function. In the exmple bove, since sine hs n odd power, we use one fctor of sin x to be the derivtive of cos x (with negtive sign), nd chnge the rest into cosine by using the pythgoren theorem: sin 3 x cos 6 x dx = (sin x)(cos 6 x)(sin x) dx = ( cos x)(cos 6 x)(sin x) dx Let u = cos x, then du = sin x dx sin x dx = du. The bove integrl becomes: ( cos x)(cos 6 x)(sin x) dx = ( u )(u 6 ) ( du) = ( u )u 6 du = ( u 6 u 8 du = 7 u7 ) 9 u9 + c = 7 cos7 x + 9 cos9 x + C If n integrl consist of even powers of sine nd cosine, we use the power reduction (hlfngle) formul to chnge ll the powers to be just liner power of cosine. Notice tht you my end up hving to pply the power reduction formul multiple times. Exmple: sin x cos x dx We use the power reduction formul to chnge both sin x nd cos x to be in terms of cos(x): [ ] [ ] sin x cos x dx = ( cos(x)) ( + cos(x)) dx = 4 ( cos(x)) ( + cos(x)) dx = cos (x) dx 4 We will use the power reduction formul one more time to reduce the power of cos (x) nd the bove becomes: cos (x) dx = 4 4 [ ( + cos(4x)) ] dx = 4 cos(4x) dx
11 = 4 cos(4x) dx = ( 4 x ) 8 sin(4x) + c = 8 x 3 sin(4x) + C We my lso use similr technique nd other trigonometric identities to integrte functions tht involve other trigonometric functions: Exmple: tn 3 x sec x dx We know tht the derivtive of sec x is sec x tn x, so if we let u = sec x, then du = sec x tn x dx will use the sec x nd one fctor of tn x. This will leve us with tn x. We cn use nother form of the pythgoren identity: tn x + = sec x to chnge between tn x nd sec x tn 3 x sec x dx = tn x (tn x sec x) dx Let u = sec x, then du = sec x tn x dx, nd tn x + = sec x tn x = sec x, the bove becomes: tn x (tn x sec x) dx = (u ) du = 3 u3 u + C = 3 sec3 x sec x + C In generl, if our integrl consist of powers of tngent nd secnt, if the power of tngent is odd, nd there is t lest one fctor of secnt, using substitution u = sec x nd chnge ll the other powers of tngent into in terms of secnt will llow us to integrte the function. Exmple: tn 5 x sec 4 x dx This integrl consist of n even power of secnts. Since the derivtive of tn x gives rise to sec x, we use the substitution: u = tn x du = sec x dx The bove becomes: tn 5 x sec 4 x dx = = (u 5 )(u + ) du = = 8 tn8 x + 6 tn6 x + C (tn 5 x)(sec x) sec x dx u 7 + u 5 du = 8 u8 + 6 u6 + C If n integrl consists of n even power of secnt nd one of more powers of tngent is present, use the substitution u = tn x nd the pproprite trigonometric
12 identities to chnge everything into in terms of u. Exmple: sec x dx This is more tricky one. Since the expression is bsent of ny tngent function, nd there is only one secnt term, none of the methods we used before will work. Insted, we introduce something extr to the integrl: sec x + tn x sec x dx = sec x sec x + tn x dx sec x + tn x Notice tht the frction is just equl to. So we did not chnge the sec x + tn x expression by multiplying this to the function sec x. However, upon multiplying this we get the integrl to new form: x + tn x (sec x)(sec x + tn x) sec sec x sec sec x + tn x dx = x + sec x tn x dx = dx sec x + tn x sec x + tn x Now we mke substitution of u = sec x + tn x, then du = (sec x tn x + sec x) dx, nd the bove integrl becomes: sec x + sec x tn x du dx = = ln u + C = ln sec x + tn x + C sec x + tn x u In other words, sec x dx = ln sec x + tn x + C Exmple: sec 3 x dx We strt with integrtion by prts. We know tht the integrl of sec x cn be esily found, so we use the substitution dv = sec x dx v = tn x u = sec x du = sec x tn x dx With the formul for by prts, we hve: sec 3 x dx = (sec x)(sec x) dx = (sec x)(tn x) (tn x)(sec x tn x) dx = sec x tn x (sec x)(tn x) dx = sec x tn x (sec x)(sec x ) dx (sec = sec x tn x 3 x sec x ) dx = sec x tn x sec 3 x dx + sec x dx Notice tht this is nother exmple where the originl integrl, fter n pplic
13 tion of by prts, shows up gin on the right, but with negtive sign. We now solve the eqution for the integrl: sec 3 x dx = sec x tn x sec 3 x dx + sec x dx sec 3 x dx = sec x tn x + sec x dx sec 3 x dx = ( sec x tn x + ) sec x dx From the previous exmple we know tht sec x dx = ln sec x + tn x + C we cn now complete the problem: sec 3 x dx = ( ) sec x tn x + sec x dx = (sec x tn x + ln sec x + tn x ) + C Exmple: tn 3 x dx We use the pythgoren identiy to chnge two of the powers of tn x into sec x then use substitution: tn 3 x dx = tn x tn x dx = (sec x )(tn x) dx = sec x tn x tn x dx = sec x tn x dx tn x dx To solve the first integrl, we mke substitution u = tn x du = sec x dx The first integrl becomes: sec x tn x dx = u du = u + C = tn x + C From the section on substitution we know tht tn x dx = ln sec x + C Putting these together gives: tn 3 x dx = sec x tn x tn x dx = sec x tn x dx tn x dx = tn x ln sec x + C
14 In generl, nytime we need to integrte n odd power of tn x, we cn turn two of the powers into sec x nd reduce the integrl to one tht involves the next odd power. A stepwise procedure like this will llow us to integrte tn n x of ny odd power. Since csc x nd cot x hve bsiclly the sme reltionship s sec x nd tn x, ny integrl tht involve csc x nd cot x cn be solved using similr technique. Trigonometric Substitution: For expressions tht involve the sum or difference of two squres, or the rdicl of sum or difference of two squres, for exmple, expressions of the form + x or x, or x...etc, one cn often use trigonometric function s substitute for the expression. It is usully esier to find the integrl of the trigonometric function thn the integrl of the originl expression. Exmple: x dx Using right tringle where the hypothenuous is, nd the opposite side to ngle θ is x, we cn turn the bove expression into one in terms of trig functions: θ x x We cn then mke the substitution ccording to the picture: x = sin θ, cos θ = x differentite we hve: dx = cos θ dθ. So the integrl becomes: x dx = cos θ cos θ dθ = = ( θ + ) sin θ + C = ( θ + cos θ dθ = ) sin θ cos θ + C ( + cos θ) dθ Notice we used the double ngle identity for sine in the bove. = (sin x + x ) x + C A trigonometric substitution my be used nytime your integrl involves the sum or difference of two squres. In other words, if you hve expressions of the form x +, or x, or x, where is constnt.
15 Depending on which form you hve, you wnt to drw the right tringle with the sides lbeled ccordingly: If the expression involves x +, you will wnt x nd to be the two legs of the right tringle. x + θ x For this cse, if you lbeled the opposite side of θ s x, you will hve to integrte function tht involves tngent nd secnt. If you lbeled the djcent side s x, you will be deling with cotngents nd cosecnts. While either of these two lbels will work, since we hve been deling with tngent nd secnt more often, it is good ide tht you lbel the opposite side x If the expression involves x, then x should be the hypotenuses, nd should be one of the two legs. x x θ For this cse, lbeling the djcent side to θ s will give you functions in terms of tngent nd secnt. Lbeling the opposite side s will give you functions in terms of cotngent nd cosecnt. If the expression involves x, then should be the hypotenuses, nd x is one of the two legs. θ x x Exmple: x 5 x + dx In this integrl, we re looking t the sum of two squres, by letting =. We wnt to put x nd on the two legs of the right tringle.
16 x + θ x According to picture, tn θ = x x = tn θ dx = sec θ dθ cos θ = x + cos θ = x + The bove expression becomes: x 5 (x x + dx = ) ( ) 5 ( ) ( 5 dx = tn θ cos θ) sec θ dθ x + = ( ( ) 5 (tn ) 5 tn 5 θ sec θ dθ = 4 θ ) (tn θ sec θ) dθ = ( ) 5 (sec θ ) (tn θ sec θ) dθ = ( ) 5 (sec 4 θ sec θ + ) (tn θ sec θ) dθ ( ) 5 [ (sec = 4 θ ) ( (tn θ sec θ) dθ sec θ ) (tn θ sec θ) dθ + ( ) ( 5 = 5 sec5 θ ) 3 sec3 θ + sec θ + C ] (tn θ sec θ) dθ Note from the picture tht x + x + x sec θ = = = + we hve bck substitution: ( ) 5 ( 5 sec5 θ 3 sec3 θ + sec θ) + C = [ ( ) ( ) 5 x 5/ 5 + ( ) x 3/ ( ) x / ] C Sme comment for trigonometric substitution s with usubstitution. When you chnge the expression in x to trigonometric function in θ, dx does not just turn into dθ. The dx term in the originl integrl must be ccounted for by differentiting both sides of the substitution eqution nd using differentils.
17 Integrtion by Prtil Frctions: Integrting rtionl expression, if the numertor hs higher degree thn the denomintor, one cn lwys perform long division to reduce it to form where we hve polynomil nd rtionl expression with the degree of numertor less thn degree of denomintor. x 3 x x + For exmple, consider: x + 3 In this rtionl function, the degree of the numertor is greter thn the degree of the denomintor. We cn perform long division of polynomils to rewrite the expression: x x + 3 ) x 3 x x + x 3 3x x 4x + x + 6 4x + 7 We cn then write the rtionl function in the form: x 3 x x + x + 3 = (x ) + 4x + 7 x + 3 We will discuss how we cn integrte rtionl function where the denomintor is qudrtic, nd the degree of the numertor is less thn the degree of the denomintor. Wht if we hve rtionl expression where the denomintor hs degree greter thn? In generl, every polynomil of rel coefficient cn be fctored into products of liner nd qudrtic terms. Wht we cn do is to fctor the donomintor of rtionl function s much s possible, then split the frction. Exmple: Evlute x + 3x 4 dx According to the method of prtil frctions, we cn fctor the denomintor nd express the rtionl function s sums of individul frctions with the fctors in the denomintor. In other words, we cn rewrite the frction like this: x + 3x 4 = (x + 4)(x ) = A x where A nd B re constnts. B x To find the vlues of A nd B, there re two methods. You my use either one:
18 First Method: If the two sides of the bove eqution re to be equl to ech other, then they must be equl for ll vlues of x. You my pick ny vlid vlues of x nd plug into the eqution. With two unknowns (A nd B), you need to pick two vlues for x. Any vlue of x tht you pick will work, s long s x 4, x. We use smll nd integrl vlues of x to mke the resulting equtions s simple s possible: Set x = 0, we hve the eqution: 4 = A 4 B Set x =, we get: 6 = A 6 + B Solving for A nd B in this system of eqution gives us: A = 5, B = 5 Second Method: Another pproch to find the vlues of A nd B is by dding bck the two frctions on the right hnd side, nd then compre coefficients of the two functions. A x B x = A(x ) + B(x + 4) (x + 4)(x ) = Ax A + Bx + 4B (x + 4)(x ) Compring the two polynomils we hve: (A + B)x + (4B A) = Equting the coefficients, we wnt A + B = 0 nd 4B A = Solving this system gives: A = 5, B = 5 = (A + B)x + (4B A) x + 3x 4 We hve: /5 x + 3x 4 dx = x /5 x dx = 5 ln x ln x + c 5 Exmple: x + x Evlute dx x 3 x After fctoring nd splitting the denomintor we get: x + x x 3 x First method: = x + x x(x )(x + ) = A x + B x + C x +
19 Setting x =, x =, nd x = 3 (your choice my be different, but the result will be the sme) gives the system of eqution: A + B + 3 C = 7 6 A 3 B C = 6 3 A + B + 4 C = 7 Solving the system of eqution gives the solution: A =, B =, C = Second method: A(x )(x + ) + B(x)(x + ) + C(x)(x ) = x(x )(x + ) = (A + B + C)x + (B C)x + ( A) x 3 x Equting coefficients gives: A + B + C = B C = = x + x x 3 x = Ax A + Bx + Bx + Cx Cx x(x )(x + ) A = Solving the system of equtions give: A =, B =, C = We hve: x + x x 3 x dx = x + x + = ln x + ln x ln x + + c x + dx In the cse if the denomintor hs nonreduceble qudrtic fctor, we need liner fctor on the numertor. Exmple: Evlute x 3 + x + x dx Fctoring gives: x 3 + x + x dx = x(x + x + ) dx
20 Notice tht the qudrtic x +x+ cnnot be further fctored by rel coefficient, we will hve: x(x + x + ) = A x + Bx + C x + x + Notice tht the numertor is liner term when the denomintor is qudrtic. First Method: We use x =, x =, nd x =. This gives, respectively, the three equtions: A + 3 B + 3 C = 3 A B + C = A + 7 B + 7 C = 7 Solving the system gives: A =, B =, C = Second Method: A x + Bx + C x + x + = A(x + x + ) + (Bx + C)(x) x(x + x + ) = Ax + Ax + A + Bx + Cx x 3 + x + x Equting coefficients give: = (A + B)x + (A + C)x + A x 3 + x + x = x 3 + x + x A + B = 0 A + C = 0 A = Solving the system gives A =, B =, C = We hve: x 3 + x + x dx = = ln x x + x x + x + dx x + x + x + + x + x + dx
21 = ln x ln x + x + x + x + dx To integrte this lst frction, we use the method of completing the squre nd use the formul for the derivtive/integrl of rctngent x + dx = ( x ) tn + c The bove integrl becomes: x + x + dx = (x + x + /4) + (3/4) dx = We now use substitution: u = x + /, the integrl becomes: ( ) u + ( 3/4) du = tn u + c 3/4 3/4 Bck substitute for x nd simplify gives: u + ( 3/4) du = ( ) x + tn + c 3 3 Finlly, putting everything together: x 3 + x + x dx = ln x ln x + x + = ln x ln x + x + 3 tn ( x + 3 ) + c (x + /) + ( 3/4) dx x + x + dx In the cse when the denomintor consist of repeted liner fctors, ech fctor nd repeted fctor needs to be ccounted for: 3x Exmple: Evlute x 3 3x + dx 3x Fctor the denomintor gives: x 3 3x + dx = 3x (x ) (x + ) dx We write: 3x (x ) (x + ) = A x + B (x ) + C x + Notice tht both of the nontrivil fctors of (x ), nmely (x ) nd (x ), re used in the prtil frction decomposition of the originl function. First Method: We hve three unknows, we need three equtions. Setting x = 0, x =, x =, gives the three equtions:
22 A + B + C = 0 A + 4 B + C = 3 4 A + B + 4 C = 3 Solving the system gives: A = /3, B =, C = /3 Second Method: A x + B (x ) + C x + = A(x + x ) + B(x + ) + C(x x + ) (x ) (x + ) = A(x )(x + ) + B(x + ) + C(x ) (x ) (x + ) = Ax + Ax A + Bx + B + Cx Cx + C (x ) (x + ) = (A + C)x + (A + B C)x + ( A + B + C) (x ) (x + ) Equting coefficients give: A + C = 0 A + B C = 3 A + B + C = 0 Solving the system of eqution gives: A = /3, B =, C = /3 The bove integrl becomes: 3x /3 x 3 3x + dx = x + (x ) /3 x + dx = 3 ln x (x ) ln x + + c 3 = 3x x 3 3x + In the cse when the denomintor hs repeted fctors of qudrtic term, ll of the fctors hve to be ccounted for. The numertor, of course, will be liner
23 term(s): Exmple: Evlute We wnt First Method: 4x (x )(x + ) dx 4x (x )(x + ) = A x + Bx + C x + + Dx + E (x + ) We hve five unknowns, we need five vlues for x, we use: x = 0, x =, x =, x =, nd x = 3 We hve the following five equtions: A + C + E = 0 A B + C 4 D + 4 E = A + 5 B + 5 C + 5 D + 5 E = A 5 B + 5 C 5 D + 5 E = 8 75 A B + 0 C D + 00 E = 3 50 Solving the system of equtions give: A =, B =, C =, D =, E =. Second Method: 4x We wnt (x )(x + ) = A x + Bx + C x + + Dx + E (x + ) = A(x + ) + (Bx + C)(x + )(x ) + (Dx + E)(x ) (x )(x + ) = A(x4 + x + ) + (Bx + C)(x 3 x + x ) + (Dx + E)(x ) (x )(x + ) (Ax 4 + Ax + A) + (Bx 4 Bx 3 + Bx Bx + Cx 3 Cx + Cx C) + (Dx Dx + Ex E) (x )(x + ) = (A + B)x4 + ( B + C)x 3 + (A + B C + D)x + ( B + C D + E)x + (A C E) (x )(x + ) Equting the coefficients we hve the following system of equtions:
24 A + B = 0 B + C = 0 A + B C + D = 0 B + C D + E = 4 A C E = 0 Solving the equtions give: A =, B =, C =, D =, E =. We now hve: 4x (x )(x + ) dx = x + x x + + x + (x + ) dx x = ln x x + dx x + dx x (x + ) dx = ln x ln x + tn x (x) (x + ) (x + ) dx = ln x ln x + tn (x) + x + + (x + ) dx To integrte the lst expression, we mke trigsubstitution: tn θ = x, then dx = sec θ dθ, cos θ =, nd the bove becomes: x + (x + ) dx = cos 4 θ sec θ dθ = cos θ dθ = ( + cos θ) dθ = (θ + ) sin(θ) + c = (θ + ) ( sin θ cos θ) + c = (θ + sin θ cos θ) + c = ( ) tn x (x) + x + + c = ( tn (x) + x ) + c x + x + Putting ll of these together: 4x (x )(x + ) dx = x + x x + + x + (x + ) dx = ln x ln x + tn (x) + x + + (x + ) dx
25 = ln x ln x + tn (x) + ( ( x + + tn (x) + = ln x ln x + tn (x) + x + + tn (x) + = ln x ln x + + x + x + + c x x + + c x )) + c x + Summry of Methods of Integrtion: As you sw from ll the integrtion methods we introduced, finding ntiderivtive is much more chllenging tsk thn tht of differentition. Unlike differentition, there is no strightforwrd formul/rule we cn use tht would solve ny integrtion problem. Anytime you encounter n integrl, you will need to consider ll the methods you hve lerned, nd see which of the method(s) you cn use to solve the problem. More often thn not, you will need to combine mny of the methods we covered in order to solve the problem. You my need to first mke substitution to chnge n expression to different form, fter which you my need to integrte by prts, or you my need to mke trigonometric substitution then integrte using trigonometric identities. In ctul problems, you will not be told which method(s) to use, it is up to you to just keep trying different things, nd use your cretivity to come up with something tht works. You my lso wnt to memorize some of the integrtion formuls tht we use often. Some of those formuls re generl enough nd will llow you to tckle the integrl of collection of expressions of certin type. Even if you do not wnt to memorize too mny integrtion formuls, t lest know tht there re integrtion tbles for the types of integrls you wnt to del with. For exmple, you my not wnt to memorize the integrtion formul for cos n x dx, but simply knowing tht there is such formul will let you know tht cos n x cn be integrted, nd you cn look up the formul when needed (nd llowed). Even with the integrtion methods we covered, there re still mny other methods of integrtion tht is not covered here. Moreover, there re continuous functions composed of elementry (polynomils, rtionl, rdicl, exponentil, logrithmic, trigonometric) functions but whose ntiderivtives cnnot be expressed s composition of elementry functions. Prctice mkes perfect. The more integrtion problems you do, the more you develop feel for the problem nd, hopefully, the more confidence you gin nd would be ble to do better.
26 Numericl Integrtion: Becuse definite integrls re so useful, we wnt to be ble to evlute definite integrl of ll kinds; but s we sw, finding ntiderivtive is not esy tsk. In the cse we re interested in the vlue of b f(x) dx, but cnnot find the ntiderivtive of f(x), we cn still pproximte the vlue of the definite integrl using numericl methods. Remember tht definite integrl, b f(x) dx is defined by Riemnn sum, so we could pproximte the definite integrl by using Riemnn sum with lrge vlues of n. We could use the left endpoint or right endpoint of the prtition in the Riemnn sum, or we could use the midpoint for the pproximtion of the definite integrl. There re lso some other formuls invented tht re populr: Midpoint Rule: If we wnted to pproximte the vlue of the definite integrl b f(x) dx, we cn divide the intervl [, b] into n equl subintervls of x, nd use the midpoint of ech subintervl, m i, s the height of the rectngle. The re of the i th rectngle will then be x(f(m i )), where m i = (x i + x i ) is the midpoint of ech intervl. We hve the Midpoint Rule for pproximting the vlue of definite integrl:
27 b f(x) dx x where x = b n n f(m i ) i= is the length of ech intervl nd m i = (x i + x i ) is the midpoint of ech intervl. In using the midpoint rules, we re pproximting the definite integrl (re) by using rectngles of equl length, nd the height of the rectngle is given by the y coordinte of the function t the midpoint of the intervl. Trpezoidl Rule: f(x) f(x ) f(x ) f(x 3 ) f(x 4 ) f(x 5 ) f(x 6 ) f(x 0 ) x 0 x x x x 3 x 4 x 5 x 6 We will pproximte definite integrl by using trpzoids insted of rectngles. Remember the re of trpzoid with bses b nd b nd height h is given by A = (b + b )h In the picture bove, the height of ech trpzoid is x, the two bses of the trpzoid is f(x i ) nd f(x i ), so the re of the ech trpzoid T i is T = [f(x 0) + f(x )][ x] T = [f(x ) + f(x )][ x]
28 T i = [f(x i ) + f(x i )][ x] The sum of ll these trpzoids is: [f(x 0 ) + f(x )][ x] + [f(x ) + f(x )][ x] + + [f(x n ) + f(x n )][ x] + [f(x n ) + f(x n )][ x] = x [f(x 0) + f(x ) + f(x ) + + f(x n ) + f(x n )] This is the Trpzoidl Rule used to pproximte the vlue of definite integrl: b f(x) dx x [f(x 0) + f(x ) + f(x ) + + f(x n ) + f(x n )] where x = b is the length of ech intervl. n Note tht in the trpezoidl rule, the coefficient of the first nd lst term of the sum (f(x 0 ) = f() nd f(x n ) = f(b) re both, nd the coefficient of ll the other terms is. In using the trpezoidl rule, we re pproximting the re using trpzoids of equl ltitude. Simpson s Rule: We my lso try to estimte definite integrl by dividing the closed intervl [, b] into equl subintervls of length x = b. This time, we use n even number n n, nd we pproximte the re of the region in ech subintervl [x i, x i+ ] by using prbol (qudrtic function).
29 f(x ) f(x ) f(x 3 ) f(x 4 ) f(x 5 ) f(x 0 ) f(x 6 ) x 0 x x x 3 x 4 x 5 x 6 In picture bove, ech x i is eqully spced. The re of the region between the intervl [x 0, x ] is pproximted with prbol which contins the three points f(x 0 ), f(x ), f(x ) (Three points uniquely define qudrtic function). The re of the region between the intervl [x, x 4 ] is pproximted with prbol tht contins the three points f(x ), f(x 3 ), f(x 4 ). We wnt to derive formul for the sum of the re of these regions. To simplify the derivtion, we consider the cse where x 0 = h, x = 0, nd x = h.
30 (0, y ) (h, y ) ( h, y 0 ) x 0 = h 0 h = x Let ( h, y 0 ), (0, y ), nd (h, y ) be the three points tht the prbol contins. Let y = Ax + Bx + C be the eqution of the prbol tht contins these three points, then the re under this prbol is given by: h h ( Ax + Bx + C ) dx Using properties of integrl nd simplify gives: h = h h ( Ax + Bx + C ) dx h ( Ax + C ) h dx + Bx dx h Notice tht Bx is n odd function, so even functions, so h h Ax + C dx = h The bove simplifies to: h h 0 Ax + C dx. h ( Ax + Bx + C ) h ( dx = Ax + C ) h dx + Bx dx h h Bx dx = 0, nd Ax nd C re both h
31 h ( = Ax + C ) ] h ) dx = [A x Cx = (A h Ch = Ah3 + CH = h ( Ah + 6C ) 3 3 We now wnt to write the expression Ah + 6C in terms of y 0, y, nd y. Since the prbol contins the three points ( h, y 0 ), (0, y ), (h, y ), we hve: y 0 = A( h) + B( h) + C = Ah Bh + C y = C 4y = 4C y = Ah + Bh + C So y 0 + 4y + y = ( Ah Bh + C ) + (4C) + (Ah + Bh + C) = Ah + 6C So the expression for the re becomes: h ( Ah + 6C ) = h 3 3 (y 0 + 4y + y ) The re of the region under the prbol is not chnged if we shift the x i to the left or right, so this formul will be vlid even if we chnge the three x coordintes to ny rbitrry x 0, x, x with the sme reltionship. In other words, the re under the prbol contining the three points (x 0, y 0 ), (x, y ), (x, y ) is h 3 (y 0 + 4y + y ). Similrly, the re under the prbol contining the three points (x, y ), (x 3, y 3 ), (x 4, y 4 ), is given by: h 3 (y + 4y 3 + y 4 ) We cn now pproximte the re, therefore the definite integrl, by the re of these regions under the prbols: b f(x) dx h 3 (y 0 + 4y + y ) + h 3 (y + 4y 3 + y 4 ) + h 3 (y 4 + 4y 5 + y 6 ) + + h 3 (y n 4 + 4y n 3 + y n ) + h 3 (y n + 4y n + y n ) = h 3 (y 0 + 4y + y + 4y 3 + y 4 + 4y 5 + y y n + 4y n + y n ) Letting f(x i ) = y i, we hve the Simpson s Rule for estimting the vlue of definite integrl: b f(x) dx x 3 [f(x 0) + 4f(x ) + f(x ) + 4f(x 3 ) + + f(x n ) + 4f(x n ) + f(x n )] where n must be n even number, x = b n Note tht the pttern in Simpson s rule goes, 4,, 4,, 4,,..., 4,, 4,. You my wnt to memorize this s the coefficient of the first nd lst terms re, then the coefficients of the inner terms lternte between 4 nd, but must strt
32 nd end with 4. In using Simpson s rule, we re pproximting the re using qudrtic functions (prbols) tht contin the midpoint nd left nd right end points of the intervl. Compred to the trpezoidl rule, Simpson s rule is more complicted to pply but it lso gives better pproximtion for the sme n.
33 Improper Integrls: Consider the integrl: x dx We cn interpret this integrl s representing the re under the curve of x from to. Notice tht for ny rel number r, the integrl: r x dx = x ] r = r is lwys less thn for ny rel number r >. This mens tht the re represented by the bove definite integrl is ctully bounded. In fct, we cn define the bove definite integrl by limit: dx = lim x t t dx = lim x t x ] t = lim t t = In generl, for ny definite integrl tht involves n upper or lower limit(s) of integrtion to be the infinity symbol (which mens tht we wnt the (net) re of the region extending indefinitely to left or right (or both) sides of the rel xis), we cn define tht s limit t infinity.
34 Definition of n Improper Integrl (of type I): If t f(x) dx is defined in the intervl [, t] for ll rel number t, we define: t f(x) dx = lim f(x) dx t If b t b f(x) dx is defined in the intervl [t, b] for ll rel number t b, we define: b f(x) dx = lim f(x) dx t t If the limit exists, we sy tht the integrl is convergent, otherwise the integrl is divergent If both f(x) dx nd f(x) dx re convergent, we cn define:
35 f(x) dx = f(x) dx + f(x) dx where could be ny rel number.
36 Exmple: Evlute 0 e x dx Ans: By definition, 0 lim t t dx = lim ex t 0 ( e t ( e 0 ) ) = 0 + = ] t t dx = lim e x dx = lim e x ex t 0 t 0 =
37 Exmple: Evlute sin ( ) x x dx Ans: By definition, sin ( ) x x t dx = lim t sin ( ) x x dx using the substitution u =, we get x t lim t = lim t cos sin ( ) x x ( t dx = lim t cos ( ) ] t x ) cos() = cos()
38 Consider this integrl: 0 ln x dx Notice tht ln x is continuous on (0, ], but is discontinuous t the lower limit of integrtion, 0. We cn define this integrl by similr method: 0 ln x dx = lim t 0 + ln x dx = lim 0 t = t t ln x dx Using the method of integrtion by prts, we get: ] ( = (0 ) ln x dx = lim t 0 + x ln x x Notice tht we used L Hospitl s Rule to evlute the limit. t lim [t t ln t] t 0 + ) = We cn use similr wy to define n improper integrl involving function over n intervl where the function hs one or more points of discontinuity. Definition of n Improper Integrl (of type II): In generl, if f is continuous on the intervl [, b) but is discontinuous on b, we define: b f(x) dx = lim t b t f(x) dx
39 Similrly, if f is continuous on (, b] but discontinuous on, we define: b f(x) dx = lim t + b t f(x) dx If there is number c, < c < b nd f is continuous on [, b] except c, nd
40 c b f(x) dx nd f(x) dx = b c c f(x) dx both exists, then we define f(x) dx + b c f(x) dx
41 Exmple: Evlute: x /3 dx Ans: Notice tht x /3 is continuous on [, ] except t x = 0, so by definition we hve: t ] t ] x /3 dx = lim x /3 dx+ lim x /3 3 3 dx = lim t 0 t 0 + t t 0 x/3 + lim t 0 + x/3 t ( 3 = lim t 0 t/3 3 ) ( 3 ( )/3 + lim t 0 + ()/3 3 ) ( (t)/3 = 0 3 ) ( ) 4 0 =
42 Exmple: Evlute 0 x dx Ans: By definition, 0 x dx = lim t 0 + t x dx = lim ln x t 0 + ] t = 0 lim ln t = 0 ( ) = t 0 + Since is not rel number, the indefinite integrl is divergent. We know tht if f(x) is n odd function nd is continuous over the intervl [, ], then the intervl. Exmple: dx is divergent since lim x t 0 not equl to 0. f(x) dx = 0. This is not true if f(x) fils to be continuous in t x dx is divergent. This improper integrl is
43 There re times when it will be difficult or even impossible to evlute n improper integrl, but it would still be helpful to know if the improper integrl is convergent or not. A theorem similr to the squeeze theorem cn be used to determine if some improper integrl is convergent or not: Comprison Test For Integrls: Suppose f nd g re continuous on [, ) nd f(x) g(x) 0 for ll x : If If f(x) dx is convergent, then g(x) dx is divergent, then g(x) dx is lso convergent. f(x) dx is lso divergent. Exmple: Determine if e x dx is convergent. Ans: It is not esy to find the ntiderivtive of e x. Insted, since e x for ll x, nd we know tht comprison test, e x dx must lso be convergent. e x e x dx is convergent (why?), so by the
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