# Math Help - Area under the curve of sin(x)

1. ## Area under the curve of sin(x)

Hi,
Please could someone help with the following problem.

(a) Sketch the curve for y = sin(x) between x = 0 and x = 2pi,

(b) Then find the total area enclosed by the curve y = sin x and the X axis between
x = 0 and x = 1.7pi

I've sketched the curve but I'm struggling with the second part, please see below.

$\int_{x1}^{x2}sin(x)dx=\int_{x1}^{x2}-cos(x)dx$

Next I integrated the area above the X axis using the limits.

$\int_{0}^{\pi}-cos(\pi)--cos\p(0))$

Then I integrated the area below the X axis using the limits.

$\int_{\pi}^{1.7\pi}-cos(1.7\pi)--cos(\pi)$

When I subtract area1 from area2 i seem to be getting an answer of 0, which doesn't seem right.
Any help appreciated.

2. ## Re: Area under the curve of sin(x)

Remember sinus function is periodic, with an specific period. Then, the area under the fuction in the first interval wich it's positive, it's the same area in the second interval wich is negative. (Try to draw it to see)

Then, when you're trying to calculate the area yo have to take the following integral:

$A=\displaystyle \int_{a}^{b}\left |f(x) \right |dx$

a,b you have to choose it propperly to get the correct answer

3. ## Re: Area under the curve of sin(x)

Maybe I'm getting confused but the area in the second interval will be smaller than the first because the original curve is sin(x) between x=0 and x=2pi. The area I have to find is sin(x) between x=0 and x=1.7pi. To find the total area enclosed I thought I would have to subtract one from the other.

I have drawn the sinewave using the following
0 time=0 degrees
pi/2(t)=90 degrees
pi(t)=180 degrees
3pi/2=270 degrees
2pi=360 degrees

4. ## Re: Area under the curve of sin(x)

Originally Posted by chrisn30
Maybe I'm getting confused but the area in the second interval will be smaller than the first because the original curve is sin(x) between x=0 and x=2pi. The area I have to find is sin(x) between x=0 and x=1.7pi. To find the total area enclosed I thought I would have to subtract one from the other.

Please, please stop using degrees. It is so out-of-date.

You want: $\int_0^\pi {\sin (x)dx} + \int_\pi ^{1.7\pi } { - \sin (x)dx}$