BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5, Problem 36RE
To determine

The value of the indefinite integral function x3x2+1dx.

Expert Solution

Answer to Problem 36RE

The value of the indefinite integral is 13x2+1(x22)+C_

Explanation of Solution

Given information:

The integral function is x3x2+1dx.

Consider (x2+1) as u.

u=x2+1 (1)

Differentiate both sides of the Equation (1).

du=2xdx (2)

Rearrange Equation (2) to find the value of x2 as shown below.

x2=u1 (3)

Substitute u for (x2+1), (u1) for x2, and 12du for xdx in the function as shown below.

x3x2+1dx=(u1)u(12du)=12(u11/21u)du=12(u1u)du (4)

The expression to find the indefinite integral value using Equation (4) as shown below.

x3x2+1dx=12(u1u)du=12(u12+1(12+1)u12+1(12+1))=12(u3/232u1/212)=u3/23u1/2 (5)

Substitute (x2+1) for u in Equation (5) as shown below.

x3x2+1dx=(x2+1)3/23(x2+1)1/2=(x2+1)1/2((x2+1)33)=13(x2+1)1/2(x22)=13x2+1(x22)+C

Therefore, the value of the indefinite integral is 13x2+1(x22)+C_.

Apply the Theorem:

The integral function f(x)dx is traditionally used for an antiderivative of f and is called an indefinite integral.

f(x)dx=F(x) (6)

Substitute x3x2+1dx for f(x) and 13x2+1(x22)+C for F(x) in Equation (6) as shown below.

x3x2+1dx=13x2+1(x22)+C (7)

Draw the graph for the function (f(x)) and its antiderivative (F(x)) using Equation (7) as shown below:

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5, Problem 36RE

Refer to Figure 1.

The function F(x) Increases with positive value of f(x) and F(x) decreases with negative value of f(x).

The minimum value of the function occurs at f(x)=0. Hence, it is reasonable from the graphical evidence.

Therefore, F(x) is an antiderivative of f(x).

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