Question

Please help! I can't figure out if i am doing it correctly!
Evaluate the below in the comments (I couldn't figure out to do it here)

Answer 1

Hint :
\[\large \sqrt{x^2} = |x|\]

Answer 2

Sorry the equation was : \[\int\limits_{0}^{\pi} (\cos^2 \theta) ^{1/2} d \theta\]

Answer 3

Like what @ganeshie8 , you have to split the integral into two, since the (1/2) power or square root means you include both positive and negative values of the function. Split up the integral.
\[\left|\int_{0}^{\pi/2} cos(x)dx\right| + \left|\int_{\pi/2}^{\pi}cos(x)dx\right|\] or you can think of it as \[\int_{0}^{\pi/2} \cos(x)dx -\int_{\pi/2}^{\pi}\cos(x)dx\]

Answer 4

I don't think i did this right? What did I do wrong?
\[\int\limits_{0}^{\pi/2} \cos (x)dx - \int\limits_{\pi/2}^{\pi} \cos (x) dx = (-\sin(\pi/2) - (-\sin (0)) - ((-\sin (\pi/2) - (-\sin(\pi)) = (-1-0) - ( -1-0) = 0??\]

Answer 5

\[\left. \sin(x)\right]_{0}^{\pi/2} = \sin(\pi/2) - 0\]\[\left. -\sin(x)\right]_{\pi/2}^{0} = -(\sin(0) - \sin(\pi/2)) = 0+\sin(\pi/2)\]\[\boxed{\sin(\pi/2) + \sin(\pi/2)}\]What does that equal?

Answer 6

Sorry I got kicked out and was just able to come back on, it equals 2, so thats the answer?

Answer 7

Yes :) good job.

Answer 8

Thank you very much! That makes complete sense now.

Answer 9

good! Glad to hear that :)