A double integral

Let T denote the triangle shown in the figure below. Use an iterated integral to evaluate

tri424×864 6.68 KB

So I set up my problem to be the graph is going from 0 to 1 in the x axis and the slope is 2-2x so my problem looks like \int_{0}^{1} \int_{0}^{2-2x} xy,dydx
Then I integrated by Y first so I got \int_{0}^{1}\frac{xy^2}{2} |_{0}^{2-2x}
Then I get \int_{0}^{1} \frac{x(2-2x)^2}{2}
→ \frac{2x^2-2x^4}{2}
→ \int_0^1 x^2-x^4
→ \frac{x^3}{3}-\frac{x^5}{5}|_{0}^{1}
→ \frac{1}{3}-\frac{1}{5}
→ \frac{5}{15}-\frac{3}{15}
=\frac{2}{15}

Set up integral as:

integrate with respect to y first. you get \frac{1}{2}xy^2|_{0}^{2-2x} then integrate with respect to x:

you get x^2-\frac{4}{3} x^3+\frac{1}{2}x^4 |_{0}^{1} which is equal to \frac{1}{6}