Science:Math Exam Resources/Courses/MATH105/April 2016/Question 02 (a)/Solution 1

Recall the fundamental theorem of calculus tells us that ${\frac {d}{dx}}\int _{0}^{x}f(t)dt=f(x)$

As we have variables at both the upper limit and lower limit of the integral , we first break the integral into two parts.

${\begin{aligned}{\frac {d}{dx}}\int _{x}^{x^{2}}\sin(t^{2})dt={\frac {d}{dx}}\int _{0}^{x^{2}}\sin(t^{2})dt+{\frac {d}{dx}}\int _{x}^{0}\sin(t^{2})dt={\color {blue}2x\sin(x^{4})-\sin(x^{2})}\end{aligned}}$

The second equality is directly followed from fundamental theorem of Calulus, because the variables are in the lower limit, we need to have a minus sign. We used chain rule to to evaluate the derivative of the first integral , let $u=x^{2},{\frac {du}{dx}}=2x$

${\frac {d}{dx}}\int _{0}^{x^{2}}\sin(t^{2})dt=2x{\frac {d}{du}}\int _{0}^{u}\sin(t^{2})dt=2x\sin(u^{2})=2x\sin(x^{4})$

Answer: $\color {blue}2x\sin(x^{4})-\sin(x^{2})$