Question

How do you find the derivative of integral of `x^2*sin(t^2)dt` from 0 to x?

Answer 1

I can answer if you meant to ask:

How do you find the derivative of integral of `t^2*sin(t^2)dt` from 0 to x?

This is a problem for the Fundamental Theorem of Calculus, Part 1, which says: If `f(x)` is defined by:

`f(x) = int_a^x g(t) dt` for `x` in some `[a,b]`, then

`g'(x) = f(x)`.

The job (one job) of this theorem is to tell us that this is the easiest question on the examination:

Find the derivative `int_0^x t^2*sin(t^2)dt`

The answer is: `x^2sin(x^2)`

Now move on to the next question.

What is the derivative of `int_3^x (t-sint)^3/(t^2+4t+4)dt`?

It is `(x-sinx)^3/(x^2+4x+4)`.

The only thing that can go wrong is you forget to change the `t`'s to `x`'s.