Worked examples · Sheet 1

Double integrals over rectangles

Theory used: Double integral over a rectangle Reduction theorem (Fubini) on rectangles Functions with separated variables

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Compute the double integral

\iint_{R} (x + 3 y^{2}) d x d y, R = [0, 1] \times [0, 2] .

\iint_{R} (x + 3 y^{2}) d x d y = \int_{0}^{1} (\int_{0}^{2} (x + 3 y^{2}) d y) d x

\int_{0}^{2} (x + 3 y^{2}) d y = [x y + y^{3}]_{y = 0}^{y = 2} = 2 x + 8

\int_{0}^{1} (2 x + 8) d x = [x^{2} + 8 x]_{0}^{1} = 1 + 8 = 9

The domain is a rectangle and the function is continuous, so Fubini’s theorem applies and the double integral reduces to two successive single integrals. In the inner integral the variable $x$ is fixed and we integrate with respect to $y$ : the antiderivative of $x + 3 y^{2}$ with respect to $y$ is $x y + y^{3}$ , therefore the inner integral equals $2 x + 8$ . The result is a function of $x$ alone, which is integrated over $[0, 1]$ and gives $9$ .

Exercise 2

Compute the double integral

\iint_{R} x e^{x y} d x d y, R = [0, 1] \times [0, 1] .

\iint_{R} x e^{x y} d x d y = \int_{0}^{1} (\int_{0}^{1} x e^{x y} d y) d x

\int_{0}^{1} x e^{x y} d y = [e^{x y}]_{y = 0}^{y = 1} = e^{x} - 1

\int_{0}^{1} (e^{x} - 1) d x = [e^{x} - x]_{0}^{1} = (e - 1) - 1 = e - 2

The two orders of integration give the same value, therefore we choose the one with the simpler computations. Integrating first with respect to $y$ , the factor $x$ is a constant and is exactly the derivative of the exponent $x y$ with respect to $y$ , so the antiderivative of $x e^{x y}$ is $e^{x y}$ and the inner integral equals $e^{x} - 1$ . Integrating first with respect to $x$ would instead require integration by parts, since $x e^{x y}$ , as a function of $x$ , is the product of a polynomial and an exponential. With the choice made, what remains is the integral of $e^{x} - 1$ over $[0, 1]$ , which equals $e - 2$ .

Exercise 3

Compute the double integral

\iint_{R} \frac{1}{( x + y ) ^{2}} d x d y, R = [0, 1] \times [1, 2] .

\iint_{R} \frac{1}{( x + y ) ^{2}} d x d y = \int_{0}^{1} (\int_{1}^{2} \frac{1}{( x + y ) ^{2}} d y) d x

\int_{1}^{2} \frac{d y}{( x + y ) ^{2}} = [- \frac{1}{x + y}]_{y = 1}^{y = 2} = \frac{1}{x + 1} - \frac{1}{x + 2}

\int_{0}^{1} (\frac{1}{x + 1} - \frac{1}{x + 2}) d x = [ln (x + 1) - ln (x + 2)]_{0}^{1} = (ln 2 - ln 3) - (ln 1 - ln 2) = 2 ln 2 - ln 3 = ln \frac{4}{3}

On $R$ the sum $x + y$ is always greater than or equal to $1$ , so the function is continuous on the rectangle and Fubini’s theorem applies. In the inner integral $x$ is a constant, therefore the antiderivative of $\frac{1}{( x + y ) ^{2}}$ with respect to $y$ is $- \frac{1}{x + y}$ . Evaluating at the limits gives the difference $\frac{1}{x + 1} - \frac{1}{x + 2}$ , which is integrated with logarithms. Collecting the terms we obtain $2 ln 2 - ln 3$ , which by the properties of logarithms is $ln \frac{4}{3}$ .