### 1 Answer

Does that mean sin^2(x) and sin^3(x) ??

Or does it mean sin(2x) and sin(3x) ?

In either case, your notation is not very good.

I’m going to assume sin^2(x) and sin^3(x), because the intersection problem is then very simple. The functions sin^2(x) and sin^3(x) intersect at x = pi/2, but on both sides of that intersection, sin^2(x) > sin^3(x), so you’d want the integral of

[sin^2(x) – sin^3(x)] dx

from x = 0 to x = pi.

The indefinite integral of sin^2(x)dx is the integral of

[1 – cos(2x)] dx/2

= x/2 – (1/4)sin(2x);

from 0 to pi, it’s pi/2.

The indefinite integral of sin^3(x)dx is the integral of

[sin(x)]*[1 – cos^2(x)] dx

= -cos(x) + (1/3)cos^3(x).

From 0 to pi, it’s

-(-1) + (1/3)(-1) – (-1) – (1/3)(1) = 4/3.

Putting this all together, you get pi/2 – 4/3

which is about 0.237.