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


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1 35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil du = F (x) dx. Then the following equlities hold: F (x) dx = F (x)+c = u + C = du, where C is n rbitrry constnt nd the lst equlity follows from the fct tht n indefinite integrl of f(u) =isu. So we cn conclude tht F (x) dx = du, provided the vribles u nd x re relted s u = F (x). This lso shows tht it is permissible to operte with dx nd du fter the integrl sign s if they were differentils. This observtion leds to net technicl trick to clculte indefinite integrls. For exmple, dx = x + ( d 2 ) x + =2 x ++C, where the substitution u =2 x + hs been used. This trick cn be generlized. Let F (u) be n indefinite integrl of continuous function f(u) on n intervl I. Let u = g(x), where g is differentible nd its rnge is the intervl I. By the chin rule, ( F (g(x))) = F (g(x))g (x) =f(g(x))g (x). In other words, F (g(x))+c is n indefinite integrl of f(g(x))g (x). On n intervl, the most generl indefinite integrl of f(u) is f(u) du = F (u)+c. Therefore, F (g(x)) nd f(u) du cn differ t most by n dditive constnt. This proves the following theorem. Theorem 42. (The Substitution Rule). Ifu = g(x) is differentible function whose rnge is n intervl I nd f is continuous on I, then (5.9) f(g(x))g (x) dx = f(g(x)) dg(x) = f(u) du.
2 35. THE SUBSTITUTION RULE 29 The substitution rule is often referred to s chnge of the integrtion vrible. It is powerful method to clculte indefinite integrls. Exmple 66. Find x sin(x 2 +)dx. Solution: x sin(x 2 +)dx = sin(x 2 +) 2 d(x2 +)= 2 sin(u) du = 2 cos(u)+c = 2 cos(x2 +)+C, where the substitution u = x 2 + hs been used. Exmple 67. Find tn(x) dx. Solution: sin(x) d(cos(x)) du tn(x) dx = cos(x) dx = = cos(x) u = ln u + C = ln cos(x) + C =ln sec(x) + C, where the substitution u = cos(x) nd the logrithm property ln(/) = ln() hve been used. The substitution rule cn be used to evlute definite integrls by mens of the fundmentl theorem of clculus. Exmple 68. Evlute 2 xex2 dx. Solution: First, find n indefinite integrl: F (x) = xe x2 dx = e x2 dx 2 = e u du = eu + C = + C. 2 ex2 where u = x 2. By the fundmentl theorem of clculus, xe x2 dx = F (2) F () = 2 (e4 ). Note tht, when evluting the integrl, the originl vrible x hs been restored in the indefinite integrl in order to pply the fundmentl theorem of clculus. The fundmentl theorem of clculus cn lso be pplied directly in the new vrible u, provided the rnge of u is properly chnged. Indeed, in the previous exmple, the nswer could hve been recovered from the indefinite integrl 2 eu + C if u = x 2 rnges from = 2 to4=2 2 s x rnges from to 2. This is especilly useful when clcultion of definite integrl requires severl chnges of the integrtion vrible.
3 3 5. INTEGRATION Theorem 43. (The Substitution Rule for Definite Integrls). Ifg is continuous on [, b] nd f is continuous on the rnge of u = g(x), then (5.2) b f(g(x))g (x) dx = g(b) g() f(u) du. Proof. Let F be n ntiderivtive of f. Then F (g(x)) is n ntiderivtive of (F (g(x))) = F (g(x))g (x) =f(g(x))g (x). By the fundmentl theorem of clculus, b f(g(x))g (x) dx = F (g(x)) b = F (g(b)) F (g()). On the other hnd, since F (u) is n ntiderivtive of f(u), the fundmentl theorem of clculus yields g(b) f(u) du = F (u) = F (g(b)) F (g()). g() g(b) g() Since the righthnd sides of these equlities coincide, so must their lefthnd sides, which implies (5.2). Exmple 69. Evlute e ln(x)/x dx. Solution: The integrnd cn be trnsformed s ln(x) dx = ln(x) d ln(x). x So the substitution u = ln(x) cn be mde. The rnge of the new integrtion vrible u is determined by the rnge of the old one: u = when x = nd u = when x = e. Thus, e ln(x) x dx = udu= u2 = Symmetry. The clcultion of definite integrl over symmetric intervl cn be simplified if the integrnd possesses symmetry properties. Theorem 44. Suppose f is continuous on symmetric intervl [, ]. Then (5.2) (5.22) 2 f(x) dx if f( x) =f(x) (f is even), if f( x) = f(x) (f is odd).
4 35. THE SUBSTITUTION RULE 3 Proof. The integrl cn be split into two integrls: ( ) f(x) dx + + f(x) dx. In the first integrl on the very righthnd side, the substitution u = x is mde so tht u = when x = nd u = when x = nd dx = du. Hence, f( u) du nd f( u) du + f(x) dx. Now, if f is even, then f( u) =f(u) nd (5.2) follows. If f is odd, then f( u) = f(u) nd (5.22) follows. The geometricl interprettion of this theorem is trnsprent. Suppose f(x) for x. The integrl A is the re under the grph of f on [,]. If f is even, then, by symmetry, the grph of f on [, ] is obtined from tht on [,] by reflection bout the y xis. Therefore, the re f(x) dx must coincide with A. If f is odd, then its grph on [, ] is obtined by the mirror reflection bout the origin so tht the re A ppers beneth the x xis. Hence, A. Exmple 7. Evlute π π sin(x3 ) dx. Solution: Unfortuntely, n ntiderivtive of sin(x 3 ) cnnot be expressed in elementry functions, nd the fundmentl theorem of clculus cnnot be used. One cn lwys evlute the integrl by tking the limit of the sequence of Riemnn sums. An lterntive solution is due to simple symmetry rgument. Note tht sin(x 3 )isnodd function, sin(( x) 3 ) = sin( x 3 )= sin(x 3 ). The integrtion intervl is lso symmetric, [ π, π]. Thus, by property (5.22), π π sin(x 3 ) dx =.
5 32 5. INTEGRATION Remrk. In the previous exmple, tke prtition of [ π, π] by points x k = k Δx, k = N, N +,...,,,,..., N,N, where Δx = π/n. Consider the Riemnn sum with smple points being the midpoints. It is then strightforwrd to show tht the Riemnn sum vnishes becuse sin(x 3 k )= sin(x 3 k ) for k =, 2,..., N.
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