To determine

To verify:

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

0≤∫-23sinx3dx≤1

Answer

0≤∫-23sinx3dx≤1

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)dx≤∫abf(x)dx≤∫abf2(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

=-sin⁡x3

=-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,

∫-23sinx3dx= ∫23sinx3 dx

Consider that 2≤x ≤3

Taking cuberoot throughout the inequality,

23 ≤ x3 ≤ 33

Using the fact 0 ≤23≈1.26,    33  (≈1.44) ≤π2 ( ≈1.57)

0 ≤ x3 ≤ π2

As the sin function is increasing on the interval 0,π2,

sin⁡0 ≤ sin⁡x3 ≤ sin⁡π2

0 ≤ sin⁡x3 ≤ 1

By using the comparison property (concept ii),

∫230 dx≤∫23sinx3dx≤∫231 dx

03-2≤∫23sinx3dx≤1(3-2)

0≤∫23sinx3dx≤1

Since

∫-23sinx3dx= ∫23sinx3dx

0≤∫-23sinx3dx≤1

Conclusion:

Therefore,

0≤∫-23sinx3dx≤1