Chapter 8.4, Problem 26E

(a)

To determine

To state: To state the reason for the given integral to be improper.

Answer to Problem 26E

It has an infinite discontinuity at the point $x=1$ .

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}}$

It is known that the integral is said to be an improper integral if the integral function becomes at any point which lies in the interval of integration.

Since the given integral has an infinite discontinuity at the point $x=1$ , the given integral is an improper integral.

(b)

To determine

To evaluate: The given improper integral or state that it diverges.

Answer to Problem 26E

The improper integral is $\frac{\pi }{2}$ , the given improper integral converges.

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}}$

Concept Used: Consider that function $f\left(x\right)$ is continuous on $\left[a,b\right)$ , then ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{c\to b^{-}}{\lim }}{\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)dx}$ .

Calculation:

The given integral has an infinite discontinuity at the point $x=1$ and continuous on $\left[0,1\right)$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}}=\underset{c\to 1^{-}}{\lim }{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}}\\ =\underset{c\to 1^{-}}{\lim }{\left[{\sin }^{-1}x\right]}_0^c\\ =\frac{\pi }{2}\end{array}$

Since the obtained result is $\frac{\pi }{2}$ , the given improper integral converges.