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Mathematics LibreTexts

5.10: 5.5E and 5.6E U-Substitution Exercises

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5.5: Substitution

In the following exercises, find the antiderivative.

261) (x+1)4dx

Answer:
15(x+1)5+C

262) (x1)5dx

263) (2x3)7dx

Answer:
112(32x)6+C

264) (3x2)11dx

265) xx2+1dx

Answer:
x2+1+C

266) x1x2dx

267) (x1)(x22x)3dx

Answer:
18(x22x)4+C

268) (x22x)(x33x2)2dx

269) cos3θdθ(Hint:cos2θ=1sin2θ)

Answer:
\displaystylesinθsin3θ3+C

270) sin3θdθ(Hint:sin2θ=1cos2θ)

271) x(1x)99dx

Solution: (1x)101101(1x)100100+C

272) t(1t2)10dt

273) (11x7)3dx

Answer:
122(711x2)+C

274) (7x11)4dx

275) cos3θsinθdθ

Answer:
cos4θ4+C
276) sin3θdθ;u=cosθ(Hint:sin2θ=1cos2θ)

(removed exercises 277- 280)

281) x2(x33)2dx

Answer:
13(x33)+C

In the following exercises, evaluate the definite integral.

292) 10x1x2dx

293) 10x1+x2dx

Answer:
u=1+x2,du=2xdx,1221u1/2du=21

294) 20t5+t2dt

295) 10t1+t3dt

Answer:
u=1+t3,du=3t2,1321u1/2du=23(21)

296) π/40sec2θtanθdθ

297) π/40sinθcos4θdθ

Answer:
u=cosθ,du=sinθdθ,11/2u4du=13(221)

J5.5.1)

J5.5.2)

5.6: Integrals Involving Exponential and Logarithmic Functions

In the following exercises, compute each indefinite integral.

320) e2xdx

321) e3xdx

Answer:
13e3x+C

322) 2xdx

323) 3xdx

Answer:
3xln3+C

324) 12xdx

325) 2xdx

Answer:
ln(x2)+C or 2ln|x|+C

326) 1x2dx

327) 1xdx

Answer:
2x+C

In the following exercises, find each indefinite integral by using appropriate substitutions.

328) lnxxdx

329) dxx(lnx)2

Answer:
1lnx+C

336) xex2dx

337) x2ex3dx

Answer:
ex33+C

338) esinxcosxdx

339) etanxsec2xdx

Answer:
etanx+C

340) elnxdxx

341) eln(1t)1tdt

Answer:
t+C

In the following exercises, evaluate the definite integral.

355) 211+2x+x23x+3x2+x3dx

Answer:
13ln(267)

356) π/40tanxdx

357) π/30sinxcosxsinx+cosxdx

Answer:
ln(31)

358) π/2π/6cscxdx

359) π/3π/4cotxdx

Answer:
12ln32

In the following exercises, integrate using the indicated substitution.

360) xx100dx;u=x100

361) y1y+1dy;u=y+1

Answer:
y2ln|y+1|+C

362) 1x23xx3dx;u=3xx3

363) sinx+cosxsinxcosxdx;u=sinxcosx

Answer:
ln|sinxcosx|+C

364) e2x1e2xdx;u=e2x

365) ln(x)1(lnx)2xdx;u=lnx

Answer:
13(1(lnx2))3/2+C

In the following exercises, f(x)0 for axb. Find the area under the graph of f(x) between the given values a and b by integrating.

372) f(x)=log10(x)x;a=10,b=100

373) f(x)=log2(x)x;a=32,b=64

Answer:
112ln2

374) f(x)=2x;a=1,b=2

375) f(x)=2x;a=3,b=4

Answer:
1ln(65,536)

376) Find the area under the graph of the function f(x)=xex2 between x=0 and x=5.

377) Compute the integral of f(x)=xex2 and find the smallest value of N such that the area under the graph f(x)=xex2 between x=N and x=N+10 is, at most, 0.01.

Answer:
N+1Nxex2dx=12(eN2e(N+1)2). The quantity is less than 0.01 when N=2.

378) Find the limit, as N tends to infinity, of the area under the graph of f(x)=xex2 between x=0 and x=5.

379) Show that badtt=1/a1/bdtt when 0<ab.

Answer:
badxx=ln(b)ln(a)=ln(1a)ln(1b)=1/a1/bdxx

380) Suppose that f(x)>0 for all x and that f and g are differentiable. Use the identity fg=eglnf and the chain rule to find the derivative of fg.

381) Use the previous exercise to find the antiderivative of h(x)=xx(1+lnx) and evaluate 32xx(1+lnx)dx.

Answer:
23

382) Show that if c>0, then the integral of 1/x from ac to bc (0<a<b) is the same as the integral of 1/x from a to b.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln(x)=x1dtt, using properties of the definite integral and making no further assumptions.

383) Use the identity ln(x)=x1dtt to derive the identity ln(1x)=lnx.

Solution: We may assume that x>1,so 1x<1. Then, 1/x1dtt. Now make the substitution u=1t, so du=dtt2 and duu=dtt, and change endpoints: 1/x1dtt=x1duu=lnx.

384) Use a change of variable in the integral xy11tdt to show that lnxy=lnx+lny for x,y>0.

385) Use the identity lnx=x1dtx to show that ln(x) is an increasing function of x on [0,), and use the previous exercises to show that the range of ln(x) is (,). Without any further assumptions, conclude that ln(x) has an inverse function defined on (,).

386) Pretend, for the moment, that we do not know that ex is the inverse function of ln(x), but keep in mind that ln(x) has an inverse function defined on (,). Call it E. Use the identity lnxy=lnx+lny to deduce that E(a+b)=E(a)E(b) for any real numbers a, b.

387) Pretend, for the moment, that we do not know that ex is the inverse function of lnx, but keep in mind that lnx has an inverse function defined on (,). Call it E. Show that E(t)=E(t).

Solution: x=E(ln(x)). Then, 1=E(lnx)x or x=E(lnx). Since any number t can be written t=lnx for some x, and for such t we have x=E(t), it follows that for any t,E(t)=E(t).

388) The sine integral, defined as S(x)=x0sinttdt is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x. Show that for k1,|S(2πk)S(2π(k+1))|1k(2k+1)π. (Hint: sin(t+π)=sint)

389) [T] The normal distribution in probability is given by p(x)=1σ2πe(xμ)2/2σ2, where σ is the standard deviation and μ is the average. The standard normal distribution in probability, ps, corresponds to μ=0 and σ=1. Compute the left endpoint estimates R10 and R100 of 1112πex2/2dx.

Solution: R10=0.6811,R100=0.6827

CNX_Calc_Figure_05_06_202.jpeg

390) [T] Compute the right endpoint estimates R50 and R100 of 53122πe(x1)2/8.


5.10: 5.5E and 5.6E U-Substitution Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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