- #1
shamieh
- 539
- 0
Evaluate the Integral.
Just wondering if someone could check my work, thanks in advance.
\(\displaystyle \int ^0_{-\infty} \frac{1}{e^{2x}} \, dx \)
\(\displaystyle lim_{a\to-\infty} \int ^0_a \frac{1}{e^{2x}} \, dx = lim_{a\to-\infty} \frac{1}{2} \int ^0_a \frac{1}{e^u}\)
*Letting \(\displaystyle u = 2x\)
&& \(\displaystyle du/2 = dx\)
\(\displaystyle
\therefore lim_{a\to-\infty} \frac{1}{2} \int ^0_a e^{-u} = lim_{a\to-\infty} \frac{1}{2} \int ^0_{2a} e^{-u}\)
\(\displaystyle = -\frac{1}{2}e^{-u} |^0_{2a}\)
\(\displaystyle = -\frac{1}{2} + \infty \)
\(\displaystyle \therefore\) as \(\displaystyle lim_{a\to-\infty}\) diverges
Just wondering if someone could check my work, thanks in advance.
\(\displaystyle \int ^0_{-\infty} \frac{1}{e^{2x}} \, dx \)
\(\displaystyle lim_{a\to-\infty} \int ^0_a \frac{1}{e^{2x}} \, dx = lim_{a\to-\infty} \frac{1}{2} \int ^0_a \frac{1}{e^u}\)
*Letting \(\displaystyle u = 2x\)
&& \(\displaystyle du/2 = dx\)
\(\displaystyle
\therefore lim_{a\to-\infty} \frac{1}{2} \int ^0_a e^{-u} = lim_{a\to-\infty} \frac{1}{2} \int ^0_{2a} e^{-u}\)
\(\displaystyle = -\frac{1}{2}e^{-u} |^0_{2a}\)
\(\displaystyle = -\frac{1}{2} + \infty \)
\(\displaystyle \therefore\) as \(\displaystyle lim_{a\to-\infty}\) diverges