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?
Best Answer
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.