Double integral with function in limit I have tried the following question.
Take the double integral of 
$$\int \int xy\,\mathrm dx\,\mathrm dy$$
$$\text{where},\,\, 0\leq x,\,\, 0\leq y, \,\,x+y \leq 1$$
I think the problem here is that I have difficulty imagening how the function looks like. By the first 2 conditions we can see that the function will be in the bottom left section of a normal $x$ and $y$ axis graph. 
First I take the integral with respect to $y$, with the upper limit of $1-x$ and the lower of 0. 
This gives me:
$$\frac{x(1-x)^2}2$$
Then I take the second derivative with respect to y with the upper limit of $0$ and the lower of $-1$.
After dong so I get $17/24$. Which is incorrect, the right answer is $1/24$. 
I did not include my calculations as im not sure how to do the integral sign. Anyway im pretty sure something is wrong with my limits/boundary points. 
Can anyone help me? Please ask if something was unclear. 
/Andreas
 A: 
The region in which you want to integrate is a triangle enclosed by the axes and the line $x+y=1$. There are two ways to do this.
1) Integrate with respect to $x$ first, then $y$. In this case, when integrating with respect to $x$, you assume you are at some fixed $y$ value. Then (using a sketch of the graph) you integrate with respect to $x$ starting at $0$ and finishing at $1-y$. Now you integrate with respect to $y$ from $0$ to $1$. So you'd get $$\int_0^1\int_0^{1-y} xy\,\mathrm dx\,\mathrm dy$$
2) Integrate with respect to $y$ first, then $x$. In this case, when integrating with respect to $y$, you assume you are at some fixed $x$ value. Then integrate with respect to $y$ starting at $0$ and finishing at $1-x$. Now you integrate with respect to $x$ from $0$ to $1$. So you'd get $$\int_0^1\int_0^{1-x} xy\,\mathrm dy\,\mathrm dx$$You'll notice that these both give the same result: $1/24$.

By the way, your error is the fact that your limits for $y$ were $-1$ to $0$. Then in the integration, you get $$\frac12\left[\frac12x^2-\frac23x^3+\frac14x^4\right]_{-1}^0=\frac12\left(\frac12\color{red}{+}\frac23+\frac14\right)=\frac{17}{24}$$
whereas the correct limits give $$\frac12\left[\frac12x^2-\frac23x^3+\frac14x^4\right]_{0}^1=\frac12\left(\frac12\color{red}{-}\frac23+\frac14\right)=\frac1{24}$$
A: First integral is right, though.  
$$\int y(1-y)^2/2 dy$$
$$=\int \dfrac{y}2-y^2+\dfrac{y^3}2 dy$$
$$=\bigg[\dfrac{y^2}4-\dfrac{y^3}3+\dfrac{y^4}8\bigg]_0^1$$
$$=\frac14-\frac13+\frac18=\frac1{24}$$
