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

Apr 28, 2015

How do you find the derivative of integral of ${t}^{2} \cdot \sin \left({t}^{2}\right) \mathrm{dt}$ from 0 to x?

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

$f \left(x\right) = {\int}_{a}^{x} g \left(t\right) \mathrm{dt}$ for $x$ in some $\left[a , b\right]$, then

$g ' \left(x\right) = f \left(x\right)$.

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

Find the derivative ${\int}_{0}^{x} {t}^{2} \cdot \sin \left({t}^{2}\right) \mathrm{dt}$

The answer is: ${x}^{2} \sin \left({x}^{2}\right)$

Now move on to the next question.

What is the derivative of ${\int}_{3}^{x} {\left(t - \sin t\right)}^{3} / \left({t}^{2} + 4 t + 4\right) \mathrm{dt}$?

It is ${\left(x - \sin x\right)}^{3} / \left({x}^{2} + 4 x + 4\right)$.

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