Method of Substitution in Calculus

8. Method ofsubstitution The chain rule says that dF(tt(x)) dx = Fj(tt(x)) ttj(x), so that Fj(tt(x)) ttj(x) dx= F(tt(x)) + C. 8.1. Example. Consider the function f (x) = 2x sin(x2 + 3). It does not appear in the list of standard antiderivatives we know by heart. But we do notice2that 2x = d(x2 + 3). So let s call tt(x) = x2 + 3, and F(u) = cosu, then dx 2 F(tt(x)) = cos(x + 3) and dF(tt(x)) dx 2 = sin(x + 3) 2x = f(x), s x t( G x) s x Ft(G(x)) so that 2xsin(x2+ 3)dx= cos(x2+ 3) + C. (58) 8.2. Leibniz notation for substitution. The most transparent way of computing an integral by substitution is by following Leibniz and introduce new variables. Thus to do the integral f(tt(x))ttj(x) dx where f (u) = F j(u), we introduce the substitution u = tt(x), and agree to write du= dtt(x) = ttj(x) dx. Then we get f(tt(x))ttj(x) dx= f(u)du= F(u) + C. At the end of the integration we must remember that u really stands for tt(x), so that f(tt(x))ttj(x) dx= F(u) + C = F(tt(x)) + C. As an example, let s do the integral (58) using Leibniz notation. We want to find 2x sin(x2 + 3) dx and decide to substitute z = x2 + 3 (the substitution variable doesn t always have to be called u). Then we compute dz= d.x2+ 3 = 2xdxand sin(x2+ 3) = sinz, so that 2xsin(x2+ 3)dx= sinzdz= cosz+ C. Finally we get rid of the substitution variable z, and we find 2xsin(x2+ 3)dx= cos.x2+ 3 + C. When we do integrals in this calculus class, we always get rid of the substitution variable because it is a variable we invented, and which does not appear in the original problem. But if you are doing an integral which appears in some longer discussion of a real-life (or real-lab) situation, then it may be that the substitution variable actually has a meaning (e.g. the effective stoichiometric modality of CQF self-inhibition ) in which 2 You will start noticing things like this after doing several examples. 2

case you may want to skip the last step and leave the integral in terms of the (meaningful) substitution variable. 8.3. Substitution for deftnite integrals. For definite integrals the chain rule d dx implies b f(tt(x))tt(x)dx= F(tt(b)) F(tt(a)). a which you can also write as b f(tt(x))tt(x)dx= x=a . j j j F(tt(x)) = F (tt(x))tt(x) = f(tt(x))tt(x) j G(b) j f(u)du. (59) u=G(a) 8.4. Example of substitution in a deftnite integral. Let s compute 1 x 1 + x2 dx, 0 using the substitution u = tt(x) = 1 + x2. Since du = 2x dx, the associated indefinite integral is 1 1+ x s x 1 u To find the definite integral you must compute the new integration bounds tt(0) and tt(1) (see equation (59).) If x runs between x = 0 and x = 1, then u = tt(x) = 1 + x2runs between u = 1 + 02= 1 and u = 1 + 12= 2, so the definite integral we must compute is 1 1 1 2 xdx= du. 2 s x 1 2du u 2 1 udu, x 1 (60) dx = 2 1 + x2 0 1 which is in our list of memorable integrals. So we find 1 2 2 1 u x 1 du = 1 ln u = 1 ln2. 1 + x2dx = 2 2 2 1 0 1 Sometimes the integrals in (60) are written as 1 2 1 x 1 x=0 1 + x2dx = u=1 udu, 2 to emphasize (and remind yourself) to which variable the bounds in the integral refer. 9. Exercises q d dq dx 1 x2 Compute thesederivatives: x .1 + t2 4 dt [Which values of q are allowed here?] 371. q d 366. 2 t dx d dt 0 2x 372. edx 1 0 d dx ln zdz 367. 373. Group Problem. x You can see the graph of the error function at http://en.wikipedia.org/wiki/Error_function (a) Compute the secondderivative of the error func- tion. How many inflection points does the graph of the error functionhave? t d dt dx 1 + x2 368. 0 1/ t d dt dx 369. 1 + x2 0 (b) The graph of the error function on Wikipedia shows that erf(x) is negative when x <0. But theerror 2x d dx 2 370. s ds x 3

3 function is defined as an integral of a positive function so it should be positive. Is Wikipedia wrong? Explain. x5 + 2 dx 395. 3 1 Compute the following indefinite integrals: . 374. 6x5 2x 4 7x dx . +3/x 5 + 4e + 7 (x 1)(3x+ 2)dx 396. 1 ( t 2/ t) dt 4 397. x x 1 375. dx , 3 8 1 3r r + 398. dr (x/a+ a/x+ xa+ ax+ ax)dx 376. 1 0 3 (x + 1) dx 399. 3 . 7 x x x + 6e + 1 dx 377. 4 1 3x2 2 4x 1 x2 + 1 dx . . 400. 1x x 378. 2 + dx 5 2 e 2x + x+ 1 4 401. dx (3x 5)dx 379. x 1 ,2 2 9 1 x+ x 4 402. dx x 2dx (hm...) 380. 4 1 1 . 5 4 4 x5+ 403. x4dx t 2dt 381. (!) 0 1 8 x 1 3x2dx 4 404. x 2dt (!!!) 382. 1 1 / 3 1 sintdt 405. (1 2x 3x2)dx 383. / 4 0 / 2 2 (cos + 2sin )d 406. 384. (5x2 4x+ 3)dx 0 1 / 2 0 (cos + sin 2 )d 407. (5y4 6y2 + 14) dy 385. 0 3 tan x cosx 1 dx 408. (y9 2y5 + 3y) dy 386. 2 / 3 0 / 2 cotx sin x 4 xdx dx 409. 387. / 3 0 36 1 + x2 dx 1 3/ 7 410. 388. x dx 1 0 , 0.5 3 1 t2 t4 1 dx 1 x2 411. 389. dt 0 1 8 2 6t t 2 (1/x)dx 412. 390. dt t4 4 1 ln 6 2 2x + 1 x 8edx 413. xdx 391. ln 3 1 9 2 t2 dt 414. (x3 1)2dx 392. 8 0 e3 1 415. u( u + 3 u) du dx 393. e2 x 0 3 2 2 |x 1|dx (x+ 1/x)2dx 416. 394. 2 1 106

2 435. Find the area of the region bounded by the parabola y2 = 4x and the line y = 2x. 2 |x x |dx 417. 1 2 436. Find the area bounded by the curve y = x(2 x)and the line x = 2y. (x 2|x|)dx 418. 1 2 437. Find the area bounded by the curve x2 = 4y and the line x = 4y 2. 2 (x |x 1|) dx 419. 0 . 438. Calculate the area of the region bounded by the parabo- las y = x2 and x = y2. 4 if 0 x <1, 2 x x5, if 1 x 2. f(x)dxwhere 420. f(x) = 0 439. Find the area of the region included between the parabola y2 = x and the line x + y = 2. f(x)dxwhere 421. . if x 0, 440. Find the area of the region bounded by the curves y= x and y = x. x, sin x, if 0 < x . f(x) = 441. Group Problem. You asked your assistant Joe to produce graphs of a function f (x), its derivative f F(x) of f(x). 422. Compute . 2 2 3 I = 2x 1 + x dx j(x) and an antiderivative 0 in two different ways: (a) Expand (1 + x2)3, multiply with 2x, and integrate each term. (b) Use the substitution u = 1 + x2. Unfortunately Joe simply labelled the graphs A, B, and C, and now he doesn t remember which graph is f , which is f which and explainyouranswer. jand which is F . Identify which graph is 423. Compute 2x.1 + x2 n dx. n I = C 424. If fj(x) = x 1/x2and f(1) = 1/2 find f(x). B x + 1 and deter- 425. Sketch the graph of the curve y = mine the area of the region enclosed by the curve, the x- axis andthe lines x = 0, x = 4. 426. Find the area under the curve y = 6x + 4 andabove the x-axis between x = 0 and x = 2. Draw a sketch of the curve. 427. Graph the curve y = 2 1 x2, x [0, 1], and find the area enclosed between the curve and the x-axis. (Don t evaluate the integral, but compare with the area under the graph of y= 1 x2.) 428. Determine the area under the curve y = a2 x2 and between the lines x = 0 and x = a. 429. Graph the curve y = 2 9 x2 and determinethe area enclosed between the curve and the x-axis. A 442. Group Problem. Below is the graph of a function y = f (x). y= f(x) c a b 430. Graph the area between the curve y2 = 4x and the line x = 3. Find the area of this region. 2 431. Find the area bounded by the curve y = 4 x and the lines y = 0 and y = 3. The function F (x) (graph not shown) is an antiderivative of f (x). Which among the following statements true? 432. Find the area enclosed between the curve y = sin 2x, 0 x /4 and the axes. (a) F(a) = F(c) (b) F (b) = 0 (c) F (b) > F(c) (d) The graph of y = F (x) has two inflection points? 433. Find the area enclosed between the curve y = cos 2x, 0 x /4 and the axes. Use a substitution to evaluate the following integrals. 434. Graph y2 + 1 = x, and find the area enclosed by the curve and the line x = 2. 5