# Integration (3)

• Aug 17th 2008, 12:56 AM
Tangera
Integration (3)
A curve is defined by the following pair of parametric equations:

$\displaystyle x = t(e^t)$ and $\displaystyle y = t sin t$, t>0. Find the area of the region enclosed by the curve, the x-axis and the line x = e.

Thank you!
• Aug 17th 2008, 02:35 AM
wingless
First plot to curve to make sure we set the integral correctly.

http://img509.imageshack.us/img509/147/gryu4.png

Now we can set the integral.

$\displaystyle \int_0^e y~dx$

$\displaystyle \int_0^e t \sin t~d(t e^t)$

$\displaystyle \int_0^1 t \sin t (e^t + t e^t)~dt$ (the bounds of integration changes because of the differential dx -> dt)

I guess you can solve from here.
• Aug 17th 2008, 07:22 AM
Tangera
Quote:

Originally Posted by wingless
$\displaystyle \int_0^1 t \sin t (e^t + t e^t)~dt$ (the bounds of integration changes because of the differential dx -> dt)

I guess you can solve from here.

How do I integrate $\displaystyle \int_0^1 t \sin t (e^t + t e^t)~dt$? Thanks for helping!
• Aug 17th 2008, 07:29 AM
wingless
Quote:

Originally Posted by Tangera
How do I integrate $\displaystyle \int_0^1 t \sin t (e^t + t e^t)~dt$? Thanks for helping!

$\displaystyle \int t \sin t (e^t + t e^t) ~dt = \int t^2 \sin t e^t ~dt + \int t \sin t e^t~dt$

Both of them has the same idea, integration by parts. For $\displaystyle \int t \sin t e^t~dt$ let $\displaystyle u = t$ and $\displaystyle dv = \sin t e^t~dt$. You'll need to find the integral of $\displaystyle \sin t e^t$ which you can find using another integration by parts.