Integrate with the definition $\int_{-\infty}^{0}e^x\sqrt{1-e^x}\,dx$ I am sorry for not providing any suggestion or info, but I really don't know where to start. I think that it could be related to Riemann Sum, but I don't have a clue about how to do this. I would really appreciate any kind of help. $$\int_{-\infty}^0 e^x\sqrt{1-e^x} \, dx$$
Change of variable $u=e^x$ leads to :

$$\int_{-\infty}^0e^x\sqrt{1-e^x}\,dx=\int_0^1\sqrt{1-u}\,du$$

This last integral is easy to compute !
If I want to give a HINT for this kind of thing, here's what I write:
$$
\int_{-\infty}^0 \sqrt{1-e^x} \Big(e^x\,dx\Big).
$$
That should suggest $du = e^x\,dx.$ Rather than write $u=e^x$ as in "Adren"'s answer, I'll go with $u=1-e^x$ so that $du = -e^x\,dx$ and the expression in the big parentheses is $-du$. Then we have
$$
\int_\text{?}^\text{?} \sqrt u \big( {-du}\big)
$$
and we need to figure out what to put where the two question marks are. When $x=0$ then $u=1-e^0=0.$ As as $x\to-\infty$ then $e^x\to0$ so $u\to1.$ We get
$$
\int_1^0 \sqrt u \big( {-du}\big) = \int_0^1 \sqrt u \, du =\cdots\cdots
$$