integrals

jeca86

Junior Member
Joined
Sep 9, 2005
Messages
62
evaluate the integral

i dont know how to draw the sign but the 3pi/2 is at the top and the 0 is at the bottom. the rest is absolute value of sinx multiplied by dx.

i got:
cos(3pi/2)-cos0=-1 is this right?
 
Hello, jeca86!

Evaluate the integral: \(\displaystyle \L\,\int^{\;\;\;\frac{3\pi}{2}}_0|\sin x|\,dx\)
I assume you know what the graph of y=sinx\displaystyle y\,=\,sin x looks like.

The graph of y=sinx\displaystyle \,y\,=\,|\sin x|\, from x=0\displaystyle x\,=\,0\, to x=3π2\displaystyle \,x\,=\,\frac{3\pi}{2}\, looks like this:
Code:
      |     ***          **
      |   *:::::*       *:| 
      | *:::::::::*   *:::|
      |*:::::::::::* *::::|
      |::::::::::::: :::::|
    --*-------------*-----+--
And you want the area under the curve . . . got it?
 
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