integration
Reverse the order of integration:
$$\int_{0}^{1}\int_{2\sqrt{x}}^{2\sqrt{x}+1} f(x,y)dydx$$.
This is my solution:
$$\int_{0}^{1}\int_{0}^{\frac{y^2}{4}}f(x,y)dxdy+\int_{1}^{2}\int_{\frac{(y-1)^2}{4}}^{\frac{y^2}{4}}f(x,y)dxdy+\int_{2}^{3}\int_{1-\frac{(y-1)^2}{4}}^{0}f(x,y)dxdy$$
But it dosen't work. What am I doing wrong? I think my third integral is bad. Could someone help me?
Here is your region:
Notice, it is bounded by the lines $y=0$ and $y=1$, so those are our bounds for $y$. Next, for any specific $y_0$, we are considering the $x$ values that range from $0$ to $y_0$, so our bounds for $x$ are $0<x<y$. Is that clear? The integral becomes:
$$\int_{0}^1\int_0^ye^{\frac{x}{y}}dxdy$$
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.
Best Answer
It helps to graph the region over which you are doing the integral. In the original case, you fix a value of $x$, and integrate over the values of $y$ between the two parabolas $y = 2\sqrt{x}$ and $y = 2\sqrt{x} +1$, and then varying $x$ from 0 to 1
Now, instead, let us fix a value of $y$. Then, we have the following cases
$$0 \leq y \leq 1$$ $$0 \leq x \leq \frac{y^2}{4}$$
This you've got right
Then,
$$1 \leq y \leq 2$$ $$\frac{(y-1)^2}{4} \leq x \leq \frac{y^2}{4}$$
This is also okay
Then,
$$2\leq y \leq 3$$ $$\frac{(y-1)^2}{4} \leq x \leq 1$$
This is where the mistake was done. The region would be bounded on the left by the upper parabola, and is not from 0 to 1 like in your attempt