Question

Evaluate the integral by reversing the order of integration

\(\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy} \)

Short answer

Therefore, the value of the double integral is\(\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy = \frac{1}{6}\left( {{e^9} - 1} \right)} \).

Step-by-step solution

Given data:

Consider the iterated integral:

\(\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy} \)

The objective is to evaluate the integral by changing the order of integration.

Using Fubini’s theorem.

Graphical representation:

Here the \(x\) limits are \(x = 3y\) to \(x = 3\). And the \(y\)limits are \(y = 0\) to \(y = 1\).

The region of integration is shown below.

Graph:

Change the order of integration:

\(\begin{array}{l}x = 3y \Rightarrow y = \frac{x}{3}\\y = 0 \Rightarrow x = 3(0) = 0\\y = 1 \Rightarrow x = 3(1) = 3\end{array}\)

The variable \(y\) varies from \(y = \frac{x}{3}\) to \(y = 1\).

And the variable \(x\) varies from \(x = 0\)to .

The new region can be written as \(D = \left\{ {(x,y)|\frac{x}{3} \le x \le 1,0 \le x \le 3} \right\}\)

Changing order of integration:

\(\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy = } \int\limits_{x = 0}^3 {\int\limits_{y = 0}^{\frac{x}{3}} {{e^{{x^2}}}} dydx} \)

The integral can be evaluated as follows:

\(\begin{array}{l}\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy = } \int\limits_{x = 0}^3 {\int\limits_{y = 0}^{\frac{x}{3}} {{e^{{x^2}}}} dydx} \\ = \int\limits_{x = 0}^3 {\left( {y{e^{{x^2}}}} \right)_{y = 0}^{y = \frac{x}{2}}dx} \\ = \int\limits_{x = 0}^3 {\left( {\frac{1}{3}x{e^{{x^2}}}} \right)dx} \\ = \int\limits_{x = 0}^3 {\left( {\frac{1}{3}{e^{{x^2}}}x} \right)dx} \end{array}\)

Use the substitution \(u = {x^2}\), \(du = 2xdx \Rightarrow xdx = \frac{{du}}{2}\) to transform the integral.

\(\begin{array}{l}\int\limits_{x = 0}^3 {\left( {\frac{1}{3}{e^{{x^2}}}x} \right)dx} = \frac{1}{3}\int\limits_{x = 0}^3 {\frac{1}{3}{e^u}du} \\ = \frac{1}{6}\int\limits_0^3 {{e^u}du} \\ = \frac{1}{6}\left( {{e^u}} \right)_0^3\\ = \frac{1}{6}\left( {{e^{{x^2}}}} \right)_0^3\\ = \frac{1}{6}\left( {{e^9} - 1} \right)\end{array}\)

Therefore, the value of the double integral is \(\int\limits_0^1 {\int\limits_{3y}^3 {{e^{{x^2}}}} dxdy = \frac{1}{6}\left( {{e^9} - 1} \right)} \).