Evaluate

**Punch** · September 13th 2010, 07:45 PM

Evaluate $\int^{\frac{\pi}{2}}_{0}2cosx+3sin2xdx + |\int^{\frac{5\pi}{6}}_{\frac{\pi}{2}}2cosx+3sin2x dx|$

The ans to this question is $4.25units^2$ which is different from my answer of $8.25units^2$

What I did

$\int^{\frac{\pi}{2}}_{0}2cosx+3sin2xdx + |\int^{\frac{5\pi}{6}}_{\frac{\pi}{2}}2cosx+3sin2x dx|$ $=$ $[2sinx-\frac{3}{2}cos2x]^{\frac{\pi}{2}}_{0}+|[2sinx-\frac{3}{2}cos2x]^{\frac{5\pi}{6}}_{\frac{\pi}{2}}|$
$=(3.5-(-1.5))+|(0.25-3.5)|$
$=8.25units^2$

**TheCoffeeMachine** · September 14th 2010, 12:32 AM

That's correct.

**Educated** · September 14th 2010, 12:36 AM

Maybe you got it right and the answer of 4.25 is wrong.

I got the answer of 8.25 as well.

Thread: Evaluate

LinkBack

Thread Tools

Display

Evaluate

Similar Math Help Forum Discussions

evaluate

evaluate

Evaluate

how do i evaluate...

Please evaluate

Search Tags