# Thread: [SOLVED] Area under a curve, confused

1. ## [SOLVED] Area under a curve, confused

Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!

2. How did you get $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$?

It should be $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{b-2}{b+2}\right) \right] -...$ according to mathcad.

3. It should be according to mathcad.
You're right, sorry about it. I made a mistake looking at my book. Anyway, I'm still stuck .

4. Originally Posted by arbolis
Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
I get the value of the integral to be $\frac{1}{4}\cdot\bigg[\lim_{b\to{\infty}}\ln\bigg[\frac{b-2}{b+2}\bigg]+\ln(5)\bigg]$

The limit converges to zero, so the integral has a value of $\frac{1}{4}\ln(5)$

Does this make sense?

--Chris

5. Originally Posted by arbolis
Find the area under this curve : $y=\frac {1}{x^2-4}$ in the region $x\geq 3$ and $y\geq 0$.
I've already checked that $y\geq 0$ is no restriction, so they simply ask me to find $\int_3^{+\infty} \frac{dx}{x^2-4}$. Looking at a formula, I found the integral's worth $\frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...$. I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
$\int_3^{\infty}\frac{dx}{x^2-4}=\int_0^{\infty}\frac{-dx}{4-x^2}=\frac{-1}{4}\int\frac{dx}{1-\left(\frac{x}{2}\right)^2}$

Now let $\varphi=\frac{x}{2}\Rightarrow{2\varphi=x}$

So $dx=2d\varphi$

So we get that

$\int\frac{dx}{x^2-4}=\frac{-1}{2}{\rm{arctanh}}\left(\frac{x}{2}\right)\bigg|_ {3}^{\infty}=\frac{\ln(5)}{4}$

6. I get the value of the integral to be

The limit converges to zero, so the integral has a value of

Does this make sense?
Yes this does make sens! I'm too tired, how could have missed it from what Plato said...
Thank you Chris.

7. If we solve the integral:

$\int_{3}^{L}\frac{1}{x^{2}-4}dx$

$=\frac{ln(5(L-2))}{4}-\frac{ln(L+2)}{4}$

Let $t=L-2$

$\lim_{L\to \infty}\left[\frac{ln(5t)}{4}-\frac{ln(t+4)}{4}\right]$

$=\frac{1}{4}\lim_{t\to {\infty}}ln(\frac{5t}{t+4})$

$=\lim_{t\to {\infty}}\frac{ln(5-\frac{20}{t+4})}{4}$

As $t\to {\infty}$, we can see we get

$\frac{ln(5)}{4}$

8. Hello, arbolis!

You did fine ... up to where you quit.

Find the area of this region: . $y\:=\:\frac {1}{x^2-4},\;x\geq 3,\;y\geq 0$
You integral is correct: . $\int^{\infty}_3\frac{dx}{x^2-4}$

The formula is usually written: . $\int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$

. . but the important feature is the absolute value.

We have: . $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3$

. $=\;
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)$

Divide top and bottom by $b$

. . $\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}$

9. Originally Posted by Soroban
Hello, arbolis!

You did fine ... up to where you quit.

You integral is correct: . $\int^{\infty}_3\frac{dx}{x^2-4}$

The formula is usually written: . $\int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$

. . but the important feature is the absolute value.

We have: . $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3$

. $=\;$ $
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" alt="
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" />

Divide top and bottom by $b$

. . $\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}$
Honestly I don't see what the big to do is over this.

$\frac{b-2}{b+2}\sim{1}\quad\text{As }b\to\infty$

$\lim_{b\to\infty}\ln\left(\frac{b-2}{b+2}\right)\sim\ln(1)=0$

Giving us

$\frac{1}{4}\bigg[0-\ln\left(\frac{1}{5}\right)\bigg]=\frac{\ln(5)}{4}$