Question

?? calculus

Answer 1

Integration, yah? c:

Answer 2

yea i already have that set up:) but i keep getting previous examples wrong:/

Answer 3

this is my last try so i decided to post it here i have 5 tries do to my homework

Answer 4

last try? :O oh boy!

Answer 5

\[\Large\rm \int\limits\limits_0^{\pi}2\sin x+\sin(2x)~dx\quad=\quad-2\cos x-\frac{1}{2}\cos(2x)~|_0^{\pi}\]

Answer 6

mm i don't have a 1/2 in my previous answer???

Answer 7

i had an x^2

Answer 8

an x^2? 0_o

Answer 9

no wait, i have that right

Answer 10

:)

Answer 11

oh ok :3

Answer 12

so where we gettin messed up?
From here it's not too bad. It's just a bunch of constants.
The cosines simply turn into 1's at the bounds.

Answer 13

Oh my bad, we do get a -1 at the upper limit, ok ok ok.
I guess we should be a LITTLE careful :)

Answer 14

\[\Large\rm =-\left[2\cos x+\frac{1}{2}\cos(2x)\right]_0^{\pi}\]
\[\Large\rm =-\left[2\cos \pi+\frac{1}{2}\cos(2\pi)\right]+\left[2\cos 0+\frac{1}{2}\cos(0)\right]\]
\[\Large\rm =-\left[-2+\frac{1}{2}\right]+\left[2+\frac{1}{2}\right]\]

Answer 15

Yah? :O

Answer 16

okay yes!!:) mine was still different before, i guess I did a little mistake as well

Answer 17

maybe im making small mistakes that are messing up my answer?

Answer 18

Different where? I haven't seen any of your work lol :3
Not sure where you're going wrong.

Answer 19

i can upload it?