# Area bounded by given curves

• Jan 30th 2010, 10:27 AM
DBA
Area bounded by given curves
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)

This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$
\int\! u\,dv = uv - \int\! v\,du
$

I have no idea how to start here. I am not even sure how the regon should look like.

Thanks for any help.
• Jan 30th 2010, 10:42 AM
e^(i*pi)
Quote:

Originally Posted by DBA
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)

This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$
\int\! u\,dv = uv - \int\! v\,du
$

I have no idea how to start here. I am not even sure how the regon should look like.

Thanks for any help.

FYI that is the integration by parts formula: Integration by parts - Wikipedia, the free encyclopedia and you're expected to use it on both equations

$5\int ln(x)$

To find the integral of ln(x) note that $\ln(x) = 1 \times \ln(x)$

Now to integrate by parts. Since we don't know the integral of ln(x) we must say that $u = \ln(x)$ and so $du = \frac{1}{x} dx$

Therefore $dv =\, dx$ giving $v = x$. Using the formula you posted

$\int \ln(x)\,dx = x\ln(x) - \int \left(x \times \frac{1}{x}\right)\, dx = x\ln(x)-x+k = x(ln(x)-1)+k$ where k is a constant

Reintroducing the 5:

$5x[ln(x)-1]+C$ where $C=5k$

Can you do the second integral? It's much the same but with x instead of 1
• Jan 30th 2010, 11:23 AM
DBA
I still do not understand what I am doing. I am looking for one area and if I use the formula on both equations I get 2 answers.
And what are the limits of the integrals?

I tried to do the second integral
Sorry I am not that good with this LaTex stuff yet, so I write an S for integral
I set u=lnx, du = 1/x dx and dv = x, v = 1/2x^2
S xln(x) = 1/2 x^2 * ln(x) - S 1/2 x^2 * 1/x dx

But I still have a product in my integral? So what do I do now?
• Jan 30th 2010, 12:03 PM
General
Quote:

Originally Posted by DBA
Find the area of the region by the given curves.
y=5ln(x) and y=xln(x)

This exercise is under the subject " Techniques of integration" and within this subject we were using the following formula
$
\int\! u\,dv = uv - \int\! v\,du
" alt="
\int\! u\,dv = uv - \int\! v\,du
" />

I have no idea how to start here. I am not even sure how the regon should look like.

Thanks for any help.

First, find the intersection points by setting the two curves equal to each other
$5ln(x)=xln(x)$
Solving, you get $x=1,5$
The area of the region = the integral from 1 to 5 for the bigger curve - the smaller curve.
To know which the is bigger, take any value for x in $(1,5)$ and substitute in both of them.
Let take $x=e$
substitute this in $y=5ln(x)$ you got 5
substitute this in $y=xln(x)$ you got e
since $5>e$
Hence, $5ln(x)>xln(x)$ for all x in $(1,5)$
So the bigger curve is $y=5ln(x)$ and the smaller curve is $y=xln(x)$

The area = $\int_1^5 ( 5ln(x) - xln(x) ) dx = 5 \int_1^5 ln(x) dx - \int_1^5 xln(x) dx$

Now, Solve this by using the integration by parts.
The final answer = the desired area.