   Chapter 5.2, Problem 23E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use the form of the definition of the integral given in Theorem 4 to evaluate the integral. ∫ − 2 0 ( x 2 + x ) d x

To determine

To evaluate: The integral function 20(x2+x)dx, using Theorem 4.

Explanation

Given information:

The integral function is 20(x2+x)dx.

Calculation:

Apply Theorem 4.

abf(x)dx=limni=1nf(xi)Δx (1)

Here, f(x) is continuous on [a,b] and Δx is width of interval.

Calculate the width Δx using relation:

Δx=ban

Substitute 0 for b and -2 for a in the relation.

Δx=ban=0(2)n=2n

Calculate xi using the relation:

xi=a+iΔx

Substitute -2 for a and 2n for Δx.

xi=2+2in

Writ the integral function using Equation (1) as follows:

abf(x)dx=20(x2+x)dx

Get f(x)=x2+x

Substitute xi for x.

f(xi)=(xi2+xi) (2)

Calculate f(xi) using the Equation (2).

Substitute (2+2in) for xi in Equation (2).

f(xi)=((2+2in)2+(2+2in))=4+4i2n2+2×(2)×(2in)2+2in=4+4i2n28in+2in2=2+4i2n26in

Calculate 20(x2+x)dx using Equation (1).

Substitute (x2+x) for f(x), -2 for a, 0 for b, (2+4i2n26in) for f(xi), and 2n for Δx in Equation (1)

abf(x)dx=limni=1nf(xi)Δx

20(x2+x)dx=limni=1n(2+4i2n26in)(2n)=limni=1n(4n+8i2n312in2)=limn[(4ni=1n1)+(8n3i=1ni2)12n2i=1ni]

20(x2+x)dx=limn((4ni=1n1)+(8n3×(12+22+

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