Science:Math Exam Resources/Courses/MATH101 B/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: $$\begin{array}{rl}A(x)& =e^{-x}\underset{x}{\overset{x^2}{\int }}{e}^{3t}\mathrm{d}\mathrm{t}\\ & ={\frac{e^{-x}}{3}(e^{3t})|}_x^{x^2}\\ & =\frac{e^{-x}}{3}(e^{3x^2}-e^{3x})\\ & =\frac{1}{3}(e^{3x^2-x}-e^{2x})\\ \end{array}$$

Finally, in order to find $A^{\prime }(x)$ , we can use chain rule to differentiate $A(x)$ term by term:

$$\begin{array}{rl}{A}^{\prime }(x)& =\frac{1}{3}(e^{3x^2-x}(3x^2-x{)}^{\prime }-e^{2x}(2x{)}^{\prime })\\ & =\frac{1}{3}(e^{3x^2-x}(6x-1)-2e^{2x}).\end{array}$$