True/Fälse: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If u=x2+1, then du=2x

(b) True or False: If dv=x2dx, then v=13x3

(c) True or False: We can apply integration by parts with u=lnxand dv=xdxto the integral lnxxdx

(d) True or False: We can apply integration by parts with u=xand dv=lnxdxto the integral lnxxdx

(e) True or False: Integration by parts has to do with reversing the product rule.

(f) True or False: Integration by parts is a good method for any integral that involves a product.

(g) True or False: In applying integration by parts, it is sometimes a good idea to choose uto be the entire integrand and let dv=dx

(h) True or False:03xexdx=xex-03exdx

Short Answer

Expert verified

Part (a) The statements is false,

Part (b) true,

Part (c) false,

Part (d) false,

Part (e) true,

Part (f) false,

Part (g) true,

Part (h) false

Step by step solution

01

Part (a) Step 1: Given information

Given the expressionu=x2+1

02

Part (a) Step 2: Integrate the expression 

We have

u=x2+1du=2xdx

So the statement is false

03

Part (b) Step 1: Given information

Given the expressiondv=x2dx

04

Part (b) Step 2:  Integrate the expression 

Integrating, we get

dv=x2dxv=x2dxv=x33

So the statement is true

05

Part (c) Step 1: Given information

Givenu=lnx

06

Part (c) Step 2: Integration by substitution

We have

u=lnx,du=1xdx

lnxxdx=udu=u22=(lnx)22=(lnx)22

So the statement is false

07

Part (d) Step 1: Given information

Given the expressionlnxxdx

08

Part (d) Step 2: Integration by substitution

We have

u=lnx,du=1xdx

lnxxdx=udu=u22=(lnx)22

So the statement is false

09

Part (e) Step 1: Given information

Given Integration by parts has to do with reversing the product rule.

10

Part (e) Step 2: Checking the reason

The statement is false. Because the integration by parts is not always a good method for any integral that involves product rule. The substitution method also sometimes can be used to solve the integrals.

11

Part (f) Step 1: Given information

Given Integration by parts is a good method for any integral that involves a product

12

Part (f) Step 2: Checking the reason

The statement is false. Because the integration by parts is not always a good method for any integral that involves product rule. The substitution method also sometimes can be used to solve the integrals.

13

Part (g) Step 1: Given information

Given In applying integration by parts, it is sometimes a good idea to choose u to be the entire integrand and letdv=dx

14

Part (g) Step 2: Checking the reason

The statement is true

15

Part (h) Step 1: Given information

Given03xexdx=xex-03exdx

16

Part (h) Step 2: Checking the reason

The statement is false

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Most popular questions from this chapter

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving1x24dx.

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving1x24dx.

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving1x2+4dx.

(d) True or False: The substitution x = 2 sin u is a suitable choice for solvingx2+45/2dx

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form x2a2.

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with x=asinu, we must consider the cases x>a and x<-a separately.

(h) True or False: When using trigonometric substitution with x=asecu, we must consider the cases x>a and x<-a separately.

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