   Chapter 4.5, Problem 52E

Chapter
Section
Textbook Problem

# Verify that f ( x ) = sin x 3 is an odd function and use that fact to show that 0 ≤ ∫ − 2 3 sin x 3 d x ≤ 1

To determine

To verify:

fx=sinx3 is an odd function and use that fact to show that

0-23sinx3dx1

Explanation

1) Concept:

i) Integrals of symmetric function

Suppose f is continuous on -a, a, if f is odd [f(-x)=-f(x)] function, then -aafxdx=0

ii) The comparison property, If f1(x)f(x)f2(x) in the interval [a, b] then

abf1(x)dxabf(x)dxabf2(x)dx

2) Given:

fx=sinx3

3) Calculation:

Given that

fx=sinx3

To show f is odd, we need to show f(-x)=-f(x),

Now f(-x)=sin-x3

=sin-x3

Since, sin-x=-sinx

=-sinx3

=-f(x)

Therefore, f(-x)=-f(x) , that is, f(x) is an odd function.

Consider the given integral,

-23sinx3dx

Splitting the interval into two parts,

= -22sinx3dx+ 23sinx3  dx

Since sinx3 is odd function, -22sinx3dx =0

= 0+ 23sinx 3 dx

So,

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