The question is:

The function f is given by $\displaystyle {f(x)}=\frac{e^\frac{1}{x}}{x^2}$ where x ≠ 0. Determine a number a<0 such that:

$\displaystyle \int_a^0 {f(x)}{dx} = {f(a)}$

What I've done so far:

We know that $\displaystyle {f'(x)}=\frac{-e^{-x}{x^2}-{2xe^{-x}}}{x^4}=\frac{{-e^{-x}}{(x+2)}}{x^3}$

And also that $\displaystyle -\int_0^a{f(x)}{dx}={f(a)}$ So using FTC1 =>

$\displaystyle \frac{d}{dx}-\int_0^a{f(x)}{dx}=\frac{d}{dx}{f(a)}$ =>

$\displaystyle -{f(a)}={-e^{-a}}\frac{a+2}{a^3}$ =>

$\displaystyle \frac{-e^{-a}}{a^2}={-e^{-a}}\frac{a+2}{a^3}$

What do I do now? Assuming I haven't made a catastrophic mistake in my working? Any tips/suggestions very much welcome.

Best,

CP