Please tell me where I went wrong.

http://img404.imageshack.us/img404/6172/int006qs6.png

http://img511.imageshack.us/img511/1...t007bi4np5.png

Thanks

Printable View

- May 7th 2007, 02:10 PMr_mathsIntegration
Please tell me where I went wrong.

http://img404.imageshack.us/img404/6172/int006qs6.png

http://img511.imageshack.us/img511/1...t007bi4np5.png

Thanks - May 7th 2007, 06:49 PMtopsquark
For starters your terminology, um..., stinks. You still have an integral sign AFTER you integrated!! I don't have the fancy fonts, but let me rewrite this.

Iave = 1/(1/300 - 0)*Int[2*sin(600t) dt, 0, 1/300]

= 300*2*Int[sin(600t) dt, 0, 1/300]

Let x = 600t ==> dx = 600 dt

Iave = 300*2*Int[sin(x) * (dx/600), 0, 2] <-- Watch your limits of integration!

= Int[sin(x) dx, 0 ,2]

= -cos(2) + cos(0)

= 1 - cos(2) = 1.41615 A

(Your given answer of 0.46 A is incorrect.)

-Dan - May 8th 2007, 05:37 AMr_maths
Thanks

I'll trust you over the book. - May 8th 2007, 05:47 AMtopsquark