Question

I need to integrate the function (6x+7y)^2 over the area of a triangle with vertices (-5,0), (0,5), (5,0).

I'll post what I tried already; it was wrong.

Answer 1

\[\int\limits_{-5}^{5} \int\limits_{0}^{5} (6x+7y)^2 dy dx\]This was wrong. I see now that this made a rectangular area.

Answer 2

\[\large \int\limits_{?}^{?}\int\limits_{R}^{?}(6x+7y)^2dA=\int\limits_{?}^{?}\int\limits_{R_1}^{?}(6x+7y)^2dA+\int\limits_{?}^{?}\int\limits_{R_2}^{?}(6x+7y)^2dA\]

Answer 3

\[\large =\int\limits_{-5}^{0}\int\limits_{0}^{(x+5)}(6x+7y)^2dy dx+\int\limits_{0}^{5}\int\limits_{0}^{(5-x)}(6x+7y)^2dy dx\]

Answer 4

Thanks very much for that! I'll try it and see what I get.
Very helpful for the 4 other questions I have to do.

Answer 5

YW

Answer 6

Here's a second solution:

Answer 7

\[\large \int\limits_{}^{}\int\limits_{R}^{}(6x+7y)^2dA=\int\limits_{}^{}\int\limits_{R}^{}(6x+7y)^2dxdy\]\[\large =\int\limits_{y=0}^{y=5}\int\limits_{x=y-5}^{x=5-y}(6x+7y)^2dxdy\]