Question

Double integral problem.
For some reason I just can't get this one. Will post in the first comment to make it neat. Please wait 1 minute.

Answer 1

\[\int\limits_{0}^{2}\int\limits_{0}^{(6-3x)/2} 6 - 3x - 2y, dydx\]

Answer 2

I need all the steps so i can see where I went wrong. Thanks.

Answer 3

Is the second limit of integration (with respect to y)
\[\frac{6-3x}{2}?\]
The text is a bit small.

Answer 4

Yes

Answer 5

Focus on the inner integral first:
\[\large\int_{0}^{(6-3x)/2}(6-3x-2y)\;dy\\
\large\left[6y-3xy-y^2\right]_{0}^{(6-3x)/2}\\
\left[6\left(\frac{6-3x}{2}\right)-3x\left(\frac{6-3x}{2}\right)-\left(\frac{6-3x}{2}\right)^2\right]-\left[6(0)-3x(0)-(0)^2\right]\\
9-9x+\frac{9}{4}x^2\]
Now the integral is
\[\int_{0}^{2}\left(9-9x+\frac{9}{4}x^2\right)\;dx\]
The integral is much simpler here, so I'll leave the rest to you.