Finding the area under a curve using integration of trig functions.

• Jun 21st 2012, 10:14 AM
Purn
Finding the area under a curve using integration of trig functions.
Hello!!

I had an integration exam today and I'm desperate to know if I answered this question correctly. As follows....

Find the area under the curve when Y= 5xsin(2x) when x=0, x=PI/2

I got 452, which I'm 99% sure is wrong but you never know!!

Chris. x
• Jun 21st 2012, 10:39 AM
tom@ballooncalculus
Re: Finding the area under a curve using integration of trig functions.
Bad luck. int 5x sin(2x) - Wolfram|Alpha

Just in case a picture helps...

http://www.ballooncalculus.org/draw/...thirtyfour.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Jun 21st 2012, 01:21 PM
Soroban
Re: Finding the area under a curve using integration of trig functions.
Hello, Purn!

Quote:

Find the area under the curve: . $y \:=\: 5x\sin(2x)\:\text{ for }x \in [0,\tfrac{\pi}{2}]$

You need to integrate by parts . . .

We have: . $I \;=\;\int 5x\sin(2x)\,dx$

. $\begin{Bmatrix}u &=& 5x && dv &=& \sin2x\,dx \\ du &=& 5\,dx && v &=& \text{-}\frac{1}{2}\cos(2x) \end{Bmatrix}$

$\text{Then: }\:I \;=\;\text{-}\tfrac{5}{2}x\cos(2x) + \tfrac{5}{2}\int\cos(2x)\,dx$

n . . . . . $=\;\text{-}\tfrac{5}{2}x\cos(2x) + \tfrac{5}{4}\sin(2x)\,\bigg]^{\frac{\pi}{2}}_0$

n . . . . . $=\;\bigg[\text{-}\tfrac{5}{2}\left(\tfrac{\pi}{2}\right)\cos\pi + \tfrac{5}{4}\sin\pi\bigg] - \bigg[\text{-}\tfrac{5}{2}(0)\cos0 + \tfrac{5}{4}\sin0\bigg]$

n . . . . . $= \; \text{-}\tfrac{5}{4}\pi(\text{-}1)$

n . . . . . $=\;\frac{5\pi}{4}$

• Jun 23rd 2012, 02:31 PM
HallsofIvy
Re: Finding the area under a curve using integration of trig functions.
Quote:

Originally Posted by Purn
Hello!!

I had an integration exam today and I'm desperate to know if I answered this question correctly. As follows....

Find the area under the curve when Y= 5xsin(2x) when x=0, x=PI/2

I got 452, which I'm 99% sure is wrong but you never know!!

You might, by the way, have notice that since sine is never larger than 1, "5x sin(x/2)", between x= 0 and $x= \pi/2$, cannot be larger than $5(\pi/2)$ and so its integral cannot be larger than $5(\pi/x)^2$ which is about 12.33. I can't imagine how you got "452"!