   Chapter 8, Problem 1PS

Chapter
Section
Textbook Problem

Wallis's Formulas(a) Evaluate the integrals ∫ − 1 1 ( 1 − x 2 ) d x   and   ∫ − 1 1 ( 1 − x 2 ) 2 d x . (b) Use Wallis’s Formulas to prove that ∫ − 1 1 ( 1 − x 2 ) n d x   = 2 2 n − 1 ( n ! ) 2 ( 2 n + 1 ) ! for all positive integers n.

(a)

To determine

To calculate: The integrals 11(1x2)dx and 11(1x2)2dx.

Explanation

Given:

The integrals:

11(1x2)dx and 11(1x2)2dx.

Formula used:

[F(x)]ab=F(b)F(a)

The power formula for integration is

xndx=xn+1n+1

The algebraic formula is

(a+b)2=a2+b2+2ab

Calculation:

Solve for 11(1x2)dx

11(1x2)dx=11dx+11(x2)dx=[xx33]11=1133(1(1)33)=23+113

Therefore,

11(1x2)dx=43

Hence, the value of the integral 11(1x2)dx is 43

(b)

To determine

To prove: 11(1x2)ndx=22n+1(n!)2(2n+1)! for all positive integers n.

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