Reverse the order of integration and evaluate the integral Reverse the order of integration and evaluate the integral:
$$\int_0^1 \int_x^\sqrt{x}{e^{x\over y}} \, dy \, dx$$
The answer is supposed to be $\frac{e}{2}-1$ but I keep getting $\frac{x}{2}(e-1)$.
This is my work:
$$\int_0^\sqrt{x} \left[e^\frac{x}{y}y\right]_0^y \, dy$$
$$\left[e^\frac{x}{y}y\right]_0^y=e^\frac{y}{y}y-e^\frac{0}{y}y=ey-y$$
$$\int_{0}^\sqrt{x}(ey-y)\,dy = \left[\frac{ey^{2}}{2}-\frac{y^{2}}{2}\right]_0^{\sqrt{x}}=\frac{ex}{2}-\frac{x}{2}=\frac{x}{2}(e-1)$$
What am I doing wrong?
 A: 
Consider the region of integration satisfied by the conditions $$0 \le x \le 1, \quad x \le y \le \sqrt{x}.$$  Integrating first with respect to $y$ and then $x$ corresponds to taking vertical strips of differential thickness $dx$, then integrating along those strips from the lower boundary corresponding to the diagonal line $y = x$ to the upper boundary parabolic arc $y = \sqrt{x}$.  If you reverse the order of integration, this clearly implies that $y$ ranges from $0$ to $1$, and $x$ should range from $x = y^2$ to $x = y$; corresponding to taking horizontal strips of differential thickness $dy$ and integrating along these strips from the lower bound of the parabolic arc to the upper bound of the diagonal line.
Hence, $$\int_{x=0}^1 \int_{y=x}^{\sqrt{x}} f(x,y) \, dy \, dx = \int_{y=0}^1 \int_{x=y^2}^y f(x,y) \, dx \, dy.$$  Now letting $f(x,y) = e^{x/y}$, the rest is a straightforward exercise.  Note that regardless of how the integration is performed, the answer cannot be a function of $x$ or $y$:  a definite integral of a function with respect to some variable of integration cannot remain a function of the variable of integration.
A: in your question, the cross scection is vertical. If you want to reverse the the order of integration, the cross section should be horizontal. so
$$\int_{0}^1\int_{x}^\sqrt{x}{e^{x\over y}}dydx=\int_{0}^1\int_{y^2}^y{e^{x\over y}}dxdy$$

A: You should have this:
$$
\int_0^1 \left( \int_x^\sqrt{x}{e^{x\over y}} \, dy \right) \, dx = \int_0^1 \left( \int_{y^2}^y e^{x/y} \, dx \right) \, dy
$$
The inside integral on the left side is tractable.
Here's how you get that:
For each fixed value of $x$ between $0$ and $1$, you have $y$ going from $x$ up to $\sqrt x$.  Draw the picture and see that that's a vertical line intersecting two curves.  Then draw a horizontal line to see what $x$ does for a fixed value of $y$.  You'll see it going from $y^2$ up to $y$.
A: To reverse the order of integration you need to think about the area your integral is being calculated on. It goes from $x$ is $0$ to $1$ and $y$ from $x$ to $\sqrt{x}$. Sketch these two curves to visualize it.

You now want to consider the range of $y$ values and then try to express the range of $x$ values as a function of $y$. In this diagram $y$ goes from $0$ to $1$. For a given $y$ value we see that $x$ goes from $y^2$ to $y$. So the integral (with reversed order) becomes:
$$\int_0^1\int_{y^2}^y e^\frac{x}{y} \, dx \, dy$$
Now evaluating the integral:
$$\int_0^1\int_{y^2}^y e^\frac{x}{y} \, dx \, dy$$
$$=\int_0^1 \left(e^\frac{x}{y}y\right)_{y^2}^y \, dy$$
$$=\int_0^1 \left(e^\frac{y}{y}y\right)-\left(e^\frac{y^2}{y}y\right) \, dy$$
$$=\int_0^1 \left(ey\right)-\left(e^yy\right)dy$$
For the second half of the integral we need to use by parts: $u=y$, $dv=e^y$ hence $du=1$, $v=e^y$.
$$=\left(\frac{ey^2}{2}\right)_0^1-\left(ye^y\right)_0^1+\int_0^1 e^y \, dy$$
$$=\frac{e}{2}-e+e-1$$
$$=\frac{e}{2}-1$$
