## Problem on reversing the order of integration

Evaluate by reversing the order of integration: $\int_0^1 \int_{x^{1/2}}^1 e^{y^3} dy \ dx$
• ## Solution

Recall that
(2) The minimum value of $y$ is $0$. The maximum value of $y$ is $1$.
(3) For a given $y$, we now find the minimum and maximum value of $x$.

For a fixed $y$, the minimum value of $x$ in the region is $x=0$. The maximum value of $x$ in the region is $x=y^2$.
Hence, we can rewrite the double integral as: $$I = \int_{y=0}^1 \int_{x=0}^{y^2} e^{y^3} dx \ dy$$
We now evaluate this double integral. The inner integral treats $y$ as fixed and varies $x$. As a result, $e^{y^3}$ is a constant and can be pulled out of the inner integral. Hence, $$I = \int_0^1 e^{y^3} \Biggl (\int_0^{y^2} dx \Biggr) \ dy$$
The inside integral evaluates to $y^2$. Thus $$I = \int_0^1 e^{y^3} y^2 dy$$
Because we see that the term outside of the exponential is very similar to the derivative of the exponent, we will solve this integral by substitution.

Let $u = y^3$. Thus, $du = 3 y^2 dy$. The lower bound of integration, $y=0$, corresponds to $u=0$. The upper bound, $y=1$ corresponds to $u=1$. Thus,
$$I = \int_0^1 e^u \frac{du}{3}.$$
Evaluating, \begin{align} I &= \frac{1}{3} \int_0^1 e^u du \\ &=\frac{1}{3} e^u \Bigr|_0^1 \\ &= \frac{1}{3}(e - 1). \end{align}