- #1
shamieh
- 538
- 0
Evaluate the Integral.
Just wondering if someone could check my work, thanks in advance.
$$\int ^0_{-\infty} \frac{1}{e^{2x}} \, dx $$
$$lim_{a\to-\infty} \int ^0_a \frac{1}{e^{2x}} \, dx = lim_{a\to-\infty} \frac{1}{2} \int ^0_a \frac{1}{e^u}$$
*Letting $$u = 2x$$
&& $$du/2 = dx$$
$$
\therefore lim_{a\to-\infty} \frac{1}{2} \int ^0_a e^{-u} = lim_{a\to-\infty} \frac{1}{2} \int ^0_{2a} e^{-u}$$
$$= -\frac{1}{2}e^{-u} |^0_{2a}$$
$$= -\frac{1}{2} + \infty $$
$$\therefore$$ as $$ lim_{a\to-\infty}$$ diverges
Just wondering if someone could check my work, thanks in advance.
$$\int ^0_{-\infty} \frac{1}{e^{2x}} \, dx $$
$$lim_{a\to-\infty} \int ^0_a \frac{1}{e^{2x}} \, dx = lim_{a\to-\infty} \frac{1}{2} \int ^0_a \frac{1}{e^u}$$
*Letting $$u = 2x$$
&& $$du/2 = dx$$
$$
\therefore lim_{a\to-\infty} \frac{1}{2} \int ^0_a e^{-u} = lim_{a\to-\infty} \frac{1}{2} \int ^0_{2a} e^{-u}$$
$$= -\frac{1}{2}e^{-u} |^0_{2a}$$
$$= -\frac{1}{2} + \infty $$
$$\therefore$$ as $$ lim_{a\to-\infty}$$ diverges