Question

If the function `f` is defined by `f(x)=int_0^x-sint^2dt` on the closed interval `-1<=x<=3`, then `f` has a local maximum at x=?

The correct answer is 2.507. But how?

Answer 1

Taking the derivative, we get

`f'(x) = -sin(x^2)`

We must now find the critical numbers.

`0 = -sin(x^2)`

`0, pi, 2pi, ... = x^2`

`x = 0, sqrt(pi), sqrt(2pi)`

Since the derivative at `x = sqrt(pi/3)` is negative, we can see that `x = sqrt(pi)` is a minimum and `x = sqrt(2pi)` is a maximum.

It follows that `x = sqrt(2pi) ~~ 2.507` as required.

Hopefully this helps!