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

1 Answer
Apr 28, 2015

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.