## Calculus: Early Transcendentals (2nd Edition)

$$\lim\limits_{a \to -\infty} \int_a^0e^xdx=1$$
To evaluate the improper integral recall that $$\int_a^\infty f(x) dx =\lim\limits_{b \to \infty} \int_a^bf(x)dx$$ So for our problem we have: $$\int_{-\infty}^0 e^x dx =\lim\limits_{a \to -\infty} \int_a^0e^xdx$$ Compute the integral: $$\lim\limits_{a \to -\infty} \int_a^0e^xdx=\lim\limits_{a \to -\infty}e^x\bigg |_a^0=\lim\limits_{a \to -\infty}\left(e^0-e^a\right)=1+0=1$$