Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 3 - Objective Problem 2

Let $f(x) = e^{-2x}e^{-x^2} = e^{-(x^2+2x)}$ and $F(x) = \displaystyle \int f(x) \ dx$ . Then we have the integral $I = \displaystyle \int_a^{a+8} f(x) \ dx = F(a+8) - F(a)$ and $I$ is maximum when $F'(a+8) - F'(a) = f(a+8) - f(a) = 0$ or

$\begin{aligned} e^{-((a+8)^2+2(a+8))} - e^{-(a^2+2a)} & = 0 \\ \implies e^{-((a+8)^2+2(a+8))} & = e^{-(a^2+2a)} \\ (a+8)^2+2(a+8) & = a^2+2a \\ a^2 + 16a+64 + 2a +16 & = a^2 + 2a \\ \implies 16a & = - 80 \\ a & = - 5 \end{aligned}$

We note that $F''(a+8) - F''(a) = f'(a+8) - f'(a) = - (2a+18)f(a+8) + (2a+2)f(a)$ . When $a=-5$ , $F''(3) - F''(-5) = - 16 f(-5) < 0$ . Therefore, $I$ is maximum when $a= \boxed{-5}$ .