Change of Order of Integration (Questions Answers)

On this page, we solve some integrals using the change of order of integration.

Solved Problems

$\boxed{\color{blue}\textbf{Q 1}:}$ Evaluate

$\displaystyle \int_{x=0}^1 \displaystyle \int_{y=0}^{\sqrt{1-x^2}} \sqrt{1-y^2} dy dx$

Answer:

We will use the change of order of integration.

Note

R:= $\begin{cases} 0 \leq y \leq \sqrt{1-x^2} \\ 0 \leq x \leq 1. \end{cases}$

So 0 ≤ y² ≤ 1-x² ⇒ 0 ≤ x²+y² ≤ 1 and 0 ≤ x ≤ 1. This region is drawn in the picture below.

Lets denote the given integral by I.

So I = $\displaystyle \int_{y=0}^1 \left( \displaystyle \int_{x=0}^{\sqrt{1-y^2}} \sqrt{1-y^2} dx\right) dy$

= $\displaystyle \int_{0}^1 \sqrt{1-y^2} \left( \displaystyle \int_{0}^{\sqrt{1-y^2}} dx\right) dy$

= $\displaystyle \int_{0}^1 \sqrt{1-y^2} \cdot \sqrt{1-y^2} dy$

= $\displaystyle \int_0^1 (1-y^2) dy$

= $\Big[ y-\dfrac{y^3}{3}\Big]_0^1$

= $(1-\dfrac{1}{3}) – (0-\dfrac{0}{3})$

= 2/3.

$\boxed{\color{blue}\textbf{Q 2}:}$ Evaluate

$\displaystyle \int_{x=0}^\pi \displaystyle \int_{y=x}^{\pi} \dfrac{\sin y}{y} dy dx$

Answer:

Here the region R is given by

R:= $\begin{cases} x \leq y \leq \pi \\ 0 \leq x \leq \pi. \end{cases}$

The region R above can also be written as follows ( vertical → horizontal).

R:= $\begin{cases} 0 \leq x \leq y \\ 0 \leq y \leq \pi. \end{cases}$

So given integral = $\displaystyle \int_{y=0}^\pi \left( \displaystyle \int_{x=0}^{y} \dfrac{\sin y}{y} dx\right) dy$

= $\displaystyle \int_{y=0}^\pi \dfrac{\sin y}{y} \left( \displaystyle \int_{x=0}^{y} dx\right) dy$

= $\displaystyle \int_{y=0}^\pi \dfrac{\sin y}{y} \cdot y dy$

= $\displaystyle \int_{y=0}^\pi \sin y dy$

= $\Big[ -\cos y\Big]_0^\pi$

= cos 0 – cos π

= 1-(-1)

= 2.

$\boxed{\color{blue}\textbf{Q 3}:}$ Evaluate

$\displaystyle \int_{0}^1 \displaystyle \int_{y}^{1} x^2 e^{xy} dx dy$

Answer:

Here the region R is given by

R:= $\begin{cases} y \leq x \leq 1 \\ 0 \leq y \leq 1. \end{cases}$

This can be rewritten as

R:= $\{(x,y) \in \mathbb{R}^2 : 0 \leq x \leq 1, 0 \leq y \leq x \}$

So I = $\displaystyle \int_{x=0}^1 \left( \displaystyle \int_{y=0}^{x} x^2 e^{xy} dy \right) dx$

= $\displaystyle \int_{x=0}^1 x^2 \left( \displaystyle \int_{0}^{x} e^{xy} dy \right) dx$

= $\displaystyle \int_{x=0}^1 x^2 \Big[ \dfrac{e^{xy}}{x}\Big]_{y=0}^x dx$

= $\displaystyle \int_{x=0}^1 x^2 \left( \dfrac{e^{x^2}}{x} – \dfrac{1}{x} \right) dx$

= $\displaystyle \int_{0}^1 x e^{x^2} dx – \displaystyle \int_{0}^1 x dx$

Put x² =z.

So xdx = $\dfrac{dz}{2}$.

x	0	1
z	0	1

∴ I = $\displaystyle \int_{0}^1 e^z \dfrac{dz}{2} – \dfrac{1}{2}$

= $\dfrac{1}{2} \Big[ e^z\Big]_0^1 – \dfrac{1}{2}$

= $\dfrac{1}{2}(e-1) – \dfrac{1}{2}$

= $\dfrac{e}{2}-1$

= (e-2)/2.

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