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

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

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

$\begin{array}{r}\frac{d}{dx}\underset{x}{\overset{x^2}{\int }}\sin (t^2)dt=\frac{d}{dx}\underset{0}{\overset{x^2}{\int }}\sin (t^2)dt+\frac{d}{dx}\underset{x}{\overset{0}{\int }}\sin (t^2)dt=2x\sin (x^4)-\sin (x^2)\end{array}$

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 $u=x^2,\frac{du}{dx}=2x$

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

Answer: $2x\sin (x^4)-\sin (x^2)$