evaluate the integral by reversing the order of integration
*please forgive me I don't know how to show integrals on here

this is a double integral
integral from 0 to 1 integral from 6y to 6 of e^x^2 dxdy
I changed the bounds to
x is between 0 and 6
y is between x/6 and 1
i get stuck on the integral between 0and 6 e^x^2(1-x/6)dx
I tried substitution but I wind up with $\displaystyle e^u/(2x)-e^u/12$ which doesn't work
How do I integrate this?

2. Originally Posted by koalamath
evaluate the integral by reversing the order of integration
*please forgive me I don't know how to show integrals on here

this is a double integral
integral from 0 to 1 integral from 6y to 6 of e^x^2 dxdy
I changed the bounds to
x is between 0 and 6
y is between x/6 and 1
i get stuck on the integral between 0and 6 e^x^2(1-x/6)dx
I tried substitution but I wind up with $\displaystyle e^u/(2x)-e^u/12$ which doesn't work
How do I integrate this?
We have by the bounds that $\displaystyle 0\leqslant{y}\leqslant{1}$ and $\displaystyle 6y\leqslant{x}\leqslant{6}$. The second inequality is equivalent to $\displaystyle y\leqslant\frac{x}{6}\leqslant{1}$. So stringing our inequalities toegether gives

$\displaystyle 0\leqslant{y}\leqslant\frac{x}{6}\leqslant{1}$

From there it is obvious that

$\displaystyle \int_0^1\int_{6y}^{6}f(x,y)dx~dy=\int_0^6\int_0^{\ frac{x}{6}}f(x,y)dy~dx$