Science:Math Exam Resources/Courses/MATH101 A/April 2025/Question 10/Solution 1

Since $e^{-x}$ does not depend on $t$, we can treat it similar to a constant and pull it out of the integral. Then, we compute the integral in $t$ as follows: $${\begin{aligned}A(x)&=e^{-x}\int _{x}^{x^{2}}e^{3t}\;{\rm {dt}}\\&=\left.{\frac {e^{-x}}{3}}(e^{3t})\right|_{x}^{x^{2}}\\&={\frac {e^{-x}}{3}}(e^{3x^{2}}-e^{3x})\\&={\frac {1}{3}}(e^{3x^{2}-x}-e^{2x})\\\end{aligned}}$$

Finally, in order to find $A'(x)$ , we can use chain rule to differentiate $A(x)$ term by term:

$${\begin{aligned}A'(x)&={\frac {1}{3}}\left(e^{3x^{2}-x}(3x^{2}-x)'-e^{2x}(2x)'\right)\\&={\frac {1}{3}}\left(e^{3x^{2}-x}(6x-1)-2e^{2x}\right).\end{aligned}}$$