# How do you use part 1 of the Fundamental Theorem of Calculus to find the derivative of y=int(sqrtt sintdt) from [x^3, sqrtx]?

Apr 18, 2015

You need one constant limit of integration to apply FTC Part 1.

So rewrite the integral:

$y = {\int}_{{x}^{3}}^{\sqrt{x}} \left(\sqrt{t} \sin t \mathrm{dt}\right) = {\int}_{{x}^{3}}^{1} \left(\sqrt{t} \sin t \mathrm{dt}\right) + {\int}_{1}^{\sqrt{x}} \left(\sqrt{t} \sin t \mathrm{dt}\right)$ ,

So $y = - {\int}_{1}^{{x}^{3}} \left(\sqrt{t} \sin t \mathrm{dt}\right) + {\int}_{1}^{\sqrt{x}} \left(\sqrt{t} \sin t \mathrm{dt}\right)$

Now use FTC 1 and the Chain Rule to get:

$- \sqrt{{x}^{3}} \sin \left({x}^{3}\right) 3 {x}^{2} + \sqrt{\sqrt{x}} \sin \left(\sqrt{x}\right) \frac{1}{2} \sqrt{x}$

$= - 3 {x}^{3} \sqrt{x} \sin \left({x}^{3}\right) + \frac{\sqrt[4]{x}}{2 \sqrt{x}} \sin \left(\sqrt{x}\right)$

Or whatever rewrite you favor as "simplified".