Chapter 5.3, Problem 63E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Using Different Methods Describe two ways to ∫ − 1 3 ( x + 2 ) d x Evaluate. Verify that each method gives the same result.

To determine

To calculate: The value of the integral 13(x+2)dx. Use two different methods.

Explanation

Given: The provided integral is,

âˆ«âˆ’13(x+2)dx

Formula used: The sum of first n natural is given by the formula,

âˆ‘i=1ni=n(n+1)2

The sum of a constant n times is written as,

âˆ‘i=1nc=nc

The area of a trapezoid with an altitude of h and parallel bases of lengths a and b is given by the formula,

(areaÂ ofÂ trapezoid)=12h(a+b)

Calculation: First method is use definite integral as a limit,

âˆ«âˆ’13(x+2)dx

The function f(x)=x+2 is integrable on the interval [âˆ’1,3] because it is continuous on [âˆ’1,3]

Moreover, the definition of integrability implies that any partition whose norm approaches 0 can be used to determine the limit. For computational convenience define Î” by subdividing [âˆ’1,3] into n subintervals of equal width,

Î”xi=3âˆ’(âˆ’1)n=4n

Choosing ci as the right endpoints of each subinterval,

ci=âˆ’1+i(4n)=4iâˆ’nn

So, the definite integral is,

âˆ«âˆ’13(x+2)dx=limâ€–Î”â€–â†’âˆžâˆ‘i=1nf(ci)â€‰Î”xi=limnâ†’âˆžâˆ‘i=1nf(ci)â€‰Î”xi

Put value of f(ci)=4iâˆ’nn+2=4i+nn and Î”xi=4n.

So,

âˆ«âˆ’13(x+2)dx=limnâ†’âˆžâˆ‘i=1n(4i+nn)(4n)

Split the sum into parts to use summation formulas,

âˆ«âˆ’13(x+2)dx=limnâ†’âˆžâˆ‘i=1n(16in2

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