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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

Let I n = 0 π / 2 sin n x   d x .

(a) Show that I 2 n + 2 I 2 n + 1 I 2 n .

(b) Use Exercise 50 to show that

I 2 n + 2 I 2 n = 2 n + 1 2 n + 2

(c) Use parts (a) and (b) to show that

2 n + 1 2 n + 2 I 2 n + 1 I 2 n 1

and deduce that lim n I 2 n + 1 / I 2 n = 1.

(d) Use part (c) and Exercises 49 and 50 to show that

lim n 2 1 2 3 4 3 4 5 6 5 6 7 2 n 2 n 1 2 n 2 n + 1 = π 2

This formula is usually written as an infinite product:

π 2 = 2 1 2 3 4 3 4 5 6 5 6 7

and is called the Wallis product.

(e) We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the figure). Find the limit of the ratios of width to height of these rectangles.

To determine

(a)

To show: the relation I2n+2I2n+1I2n

Explanation

A reduction formula can be obtained by using the technique of integration by parts to solve the integral sinnxdx.

Formula used:

Reduction formula is as given below:

sinnxdx=1ncosxsinn1x+n1nsinn2xdx

Given:

The relation to prove and the integral, In=0π2sinnxdx.

Calculation:

Apply the limits of integration in the reduction formula:

0π2sinnxdx=1ncosxsinn1x]0π2+n1n0π2sinn2xdx=1ncosπ2sinn1π2(1ncos0sinn10)+n1n0π2sinn2xdx

Recall that cosπ2=0 and sin0=0. So, the above integral will become:

0π2sinnxdx=1n(0)sinn1π2(1ncos0(0))+n1n0π2sinn2xdx=n1n0π2sinn2xdx

Solve the integral 0π2sin2n+1xdx using the above formula with n as 2n+1. So, substitute for n as 2n+1

0π2sin2n+1xdx=2n+112n+10π2sin2n+12xdx=2n2n+10π2sin2n1xdx …… (1)

Solve the integral 0π2sin2n1xdx again using the formula with n as 2n1

0π2sin2n1xdx=2n112n10π2sin2n12xdx=2n22n10π2sin2n3xdx

Substitute the result for integration in equation (1)

0π2sin2n+1xdx=2n+112n+10π2sin2n+12xdx=2n2n+12n22n10

To determine

(b)

To show: the relation I2n+2I2n=2n+12n+2

To determine

(c)

To show: the relation 2n+12n+2I2n+1I2n1 and limnI2n+2I2n+1=1

To determine

(d)

To show: the relation limn2123434565672n2n12n2n+1=π2

To determine

(e)

To find: the limit of ratios of the width to height of the rectangles

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