Question

Calculus
For the following determine if the integral converges. If the integral converges, compute its value.
1∫inf x/(x^2+3) dx

Answer 1

\[\int_1^\infty\frac{x}{x^2+3}~dx=\lim_{t\to\infty}\int_1^t\frac{x}{x^2+3}~dx\]
Substitute \(u=x^2+3\), then \(\dfrac{1}{2}du=x~dx\). Remember to change the limits:
\[\lim_{t\to\infty}\int_1^t\frac{x}{x^2+3}~dx=\frac{1}{2}\lim_{t\to\infty}\int_4^{t^2+3}\frac{du}{u}\]

Answer 2

So it diverges?

Answer 3

\[\begin{align*}\frac{1}{2}\lim_{t\to\infty}\int_4^{t^2+3}\frac{du}{u}&=\frac{1}{2}\lim_{t\to\infty}\bigg[\ln u\bigg]_4^{t^2+3}\\
&=\frac{1}{2}\lim_{t\to\infty}\bigg[\ln (t^2+3)-\ln4\bigg]\\
&=\frac{1}{2}\color{red}{\lim_{t\to\infty}\ln (t^2+3)}-\frac{1}{2}\ln4\end{align*}\]
What do you think? Does this function approach a finite value or infinity?

Answer 4

Looks like it will approach 0-(1/2)ln4 which is approximately .693147

Answer 5

Why does it approach 0?

Answer 6

As \(t\to\infty\), you have \(t^2+3\to\infty\) as well. which means \(\ln(t^2+3)\to\cdots\)

Answer 7

approaches infinity

Answer 8

Right, so the integral diverges

Answer 9

ok great!

Answer 10

what about 1∫inf xe^(-x^2)?

Answer 11

Substituting \(u=x^2\), you get \(\dfrac{1}{2}du=x~dx\), so
\[\int_1^\infty xe^{-x^2}~dx=\frac{1}{2}\int_1^\infty\frac{du}{e^{u}}=-\frac{1}{2}\bigg[\frac{1}{e^u}\bigg]_1^\infty=-\frac{1}{2}\lim_{t\to\infty}\frac{1}{e^t}+\frac{1}{2e}=\cdots\]

Answer 12

=1/2e