Question

can someone show me the solution for \[\int_{-3}^3 \int_0^{\pi / 2} (y + y^2 \cos x) dx dy\]

i got zero but the book says 18 >.<

Answer 1

\[\Large \int_{-3}^3 \int_0^{\pi / 2} (y + y^2 \cos x)dxdy\] i just wanna latex :p

Answer 2

\[\int\limits_{-3}^3[ \int\limits_0^{\pi / 2} (y + y^2 \cos x) dx] dy\]
integrate the bracket inside first

Answer 3

i believe i got \[\Large \frac{\pi y}{2} - y^2\]

Answer 4

\[\int\limits \left(y+y^2 \cos (x)\right) \, dx\]
i get\[ \frac{1}{2} y (2 y+\pi )\]

Answer 5

oh wait...i got \[\large [yx|_0^{\pi /2} - [y^2 \sin x|_0^{\pi /2}\]

Answer 6

yeah just insert the pi/2 into x

Answer 7

*facepalm*

Answer 8

it's supposed to be plus darn =_=

Answer 9

but im still getting 0 >.<

Answer 10

oh wait nevermind...i saw it

Answer 11

\[y^2\int\limits \cos (x) \, dx+\int\limits y \, dx\]
\[y^2 \sin (x)+\int\limits y \, dx\]
Integrated inside the brackets i get
\[[y^2 \sin (x)+x y]_{0}^{pi/2}\]

Answer 12

ok

Answer 13

seems my mistake lied on \[\frac{3^3}{3} - (-\frac{3^3}{3})\] which then becomes positive