We have $$ f(x) := \frac{e^{1/x}}{x^2}, \qquad x \ne 0$$
We need to determine a number $a<0$ such that $$ \int^0_a f(x)\, dx = f(a). $$
What I tried:
With the substitution technique I get to $[-e^{1/x}]^0_a$. So that means we would get $-e^{1/0} + e^{1/a}$, but $1/0$ is undefined, so how do I continue?
Note that your function $f$ is not defined at 0, hence you may not apply the fundamental theorem on $[a,0]$. But if you recall that $$ \int_a^0 f(x) \, dx = \lim_{b \nearrow 0} \int_a^b f(x)\, dx $$ you get, by the fundamental theorem $$ \int_a^0 f(x)\, dx = \lim_{b\nearrow 0} -e^{1/b} + e^{1/a} = e^{1/a} $$ I'm sure, you can continue from here.
