Worked examples · Sheet 3

Changing the order of integration

Theory used: Reduction formulas on normal domains Changing the order of integration

Exercise 1

Compute the iterated integral

\int_{0}^{1} (\int_{x}^{1} e^{y^{2}} d y) d x .

D = {(x, y) : 0 \leq x \leq 1, x \leq y \leq 1} = {(x, y) : 0 \leq y \leq 1, 0 \leq x \leq y}

\int_{0}^{1} (\int_{x}^{1} e^{y^{2}} d y) d x = \int_{0}^{1} (\int_{0}^{y} e^{y^{2}} d x) d y = \int_{0}^{1} y e^{y^{2}} d y

\int_{0}^{1} y e^{y^{2}} d y = [\frac{e ^{y^{2}}}{2}]_{0}^{1} = \frac{e - 1}{2}

The function $e^{y^{2}}$ does not have an elementary antiderivative with respect to $y$ , so the given inner integral cannot be computed directly and the only way forward is changing the order of integration. The given limits describe the triangle lying between the line $y = x$ , the line $y = 1$ and the $y$ -axis: the same triangle, described with respect to the $y$ -axis, has $y \in [0, 1]$ and, for each fixed $y$ , $x \in [0, y]$ . With the new order the function does not depend on $x$ , therefore the inner integral equals $y e^{y^{2}}$ . The factor $y$ is, up to the constant $2$ , the derivative of the exponent $y^{2}$ , so the antiderivative is $\frac{e ^{y^{2}}}{2}$ and the result is $\frac{e - 1}{2}$ .

Exercise 2

Compute the iterated integral

\int_{0}^{1} (\int_{x}^{1} sin (y^{3}) d y) d x .

D = {(x, y) : 0 \leq x \leq 1, x \leq y \leq 1} = {(x, y) : 0 \leq y \leq 1, 0 \leq x \leq y^{2}}

\int_{0}^{1} (\int_{x}^{1} sin (y^{3}) d y) d x = \int_{0}^{1} (\int_{0}^{y^{2}} sin (y^{3}) d x) d y = \int_{0}^{1} y^{2} sin (y^{3}) d y

\int_{0}^{1} y^{2} sin (y^{3}) d y = [- \frac{cos ( y ^{3} )}{3}]_{0}^{1} = \frac{1 - cos 1}{3}

y = \sqrt{x}

The region

0 \leq x \leq 1

x \leq y \leq 1

: the horizontal segment at height

y

goes from

x = 0

x = y^{2}

In this case too the function $sin (y^{3})$ has no elementary antiderivative, so we change the order of integration. The condition $x \leq y$ with $x, y \geq 0$ is equivalent to $x \leq y^{2}$ , therefore the region is the one lying between the $y$ -axis, the line $y = 1$ and the parabola $x = y^{2}$ : with respect to the $y$ -axis we have $y \in [0, 1]$ and $x \in [0, y^{2}]$ . The inner integral in the new order equals $y^{2} sin (y^{3})$ , and the factor $y^{2}$ is, up to the constant $3$ , the derivative of $y^{3}$ . The antiderivative is therefore $- \frac{cos ( y ^{3} )}{3}$ and the result is $\frac{1 - cos 1}{3}$ .

Exercise 3

Compute the iterated integral

\int_{0}^{1} (\int_{y}^{1} \frac{sin x}{x} d x) d y .

D = {(x, y) : 0 \leq y \leq 1, y \leq x \leq 1} = {(x, y) : 0 \leq x \leq 1, 0 \leq y \leq x}

\int_{0}^{1} (\int_{y}^{1} \frac{sin x}{x} d x) d y = \int_{0}^{1} (\int_{0}^{x} \frac{sin x}{x} d y) d x = \int_{0}^{1} \frac{sin x}{x} \cdot x d x

\int_{0}^{1} sin x d x = [- cos x]_{0}^{1} = 1 - cos 1

The function $\frac{sin x}{x}$ has no elementary antiderivative, so the given inner integral, which is in $x$ , cannot be computed directly. The region described by the limits is the triangle bounded by the lines $x = y$ , $x = 1$ and by the $x$ -axis; described with respect to the $x$ -axis, it has $x \in [0, 1]$ and $y \in [0, x]$ . With the new order the function does not depend on $y$ , therefore the inner integral equals $\frac{sin x}{x}$ multiplied by the length $x$ of the vertical segment. The product simplifies to $sin x$ , whose integral over $[0, 1]$ equals $1 - cos 1$ .