Evaluating Double Integrals over Rectangles Examples 1

# Evaluating Double Integrals over Rectangles Examples 1

Recall from the Fubini's Theorem and Evaluating Double Integrals over Rectangles page that if $z = f(x, y)$ is a two variable real-valued function and $f$ is continuous on the rectangle $R = [a, b] \times [c, d]$, then:

(1)
\begin{align} \quad \iint_R f(x, y) \: dA = \int_a^b \int_c^d f(x, y) \: dy \: dx = \int_c^d \int_a^b f(x, y) \: dx \: dy \end{align}

Furthermore, if $f$ can be written as a product of the single variable functions $g(x)$ and $h(y)$, that is $f(x, y) = g(x) h(y)$, then:

(2)
\begin{align} \quad \iint_R f(x, y) \: dA = \left [ \int_a^b g(x) \: dx \right ] \left [ \int_c^d h(y) \: dy \right ] \end{align}

We will now look at some examples of evaluating double integrals over rectangles. More examples can be found on the Evaluating Double Integrals over Rectangles Examples 2 page.

## Example 1

Evaluate the double integral $\iint_R \frac{xe^x}{y} \: dy \: dx$ where $R = [0, 1] \times [1, 2]$.

We can set the double integral above as an iterated integral as follows:

(3)
\begin{align} \quad \iint_R \frac{xe^x}{y} \: dy \: dx = \int_0^1 \int_1^2 \frac{xe^x}{y} \: dy \: dx \end{align}

Let's first evaluate the inner integral $\int_1^2 \frac{xe^x}{y} \: dy$ holding $x$ as fixed

(4)
\begin{align} \quad \int_1^2 \frac{xe^x}{y} \: dy = \left [ xe^x \ln \mid y \mid \right ]_1^2 = \ln (2) xe^x \end{align}

Therefore we have that:

(5)
\begin{align} \quad \iint_R \frac{xe^x}{y} \: dy \: dx = \int_0^1 \ln (2) xe^x \: dx = \ln(2) \left [ xe^x - e^x \right ]_0^1 = \ln (2) \end{align}

Note that $\int xe^x \: dx = xe^x - e^x$ can be obtained by using either Integration by Parts or Tabular Integration.

## Example 2

Evaluate the double integral $\iint_R \cos (x - y) \: dA$ where $R = \left [ 0 , \frac{\pi}{2} \right ] \times \left [0, \frac{\pi}{2} \right ]$.

We can rewrite this double integral as an iterated integral:

(6)
\begin{align} \quad \iint_R \cos (x - y) \: dA = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \cos (x - y) \: dx \: dy \end{align}

We will first evaluate the inner integral $\int_0^{\frac{\pi}{2}} \cos (x - y) \: dx$ by holding $y$ as fixed as follows:

(7)
\begin{align} \quad \int_0^{\frac{\pi}{2}} \cos (x - y) \: dx = \left [ \sin (x - y) \right ]_0^{\frac{\pi}{2}} = \sin \left ( \frac{\pi}{2} - y \right ) - \sin (-y) \end{align}

Therefore we have that:

(8)
\begin{align} \quad \iint_R \cos (x - y) \: dA = \int_0^{\frac{\pi}{2}} \sin \left ( \frac{\pi}{2} - y \right ) - \sin (-y) \: dy = \left [ \cos \left ( \frac{\pi}{2} - y \right ) - \cos (-y) \right ]_0^{\frac{\pi}{2}} = 1 + 1 = 2 \end{align}

## Example 3

Evaluate the double integral $\iint_R x^2 y^3 \: dA$ where $R = \left [ 1, 2 \right ] \times \left [1,3 \right ]$.

This integral can be rewritten and evaluated as:

(9)
\begin{align} \quad \iint_R x^2 y^3 \: dA = \left [ \int_1^2 x^2 \: dx \right ] \left [ \int_1^3 y^3 \: dy \right ] = \left [ \frac{x^3}{3} \right ]_1^2 \cdot \left [ \frac{y^4}{4} \right ]_1^3 = \left [ \frac{7}{3} \right ] \left [ 20\right] = \frac{140}{3} \end{align}
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