Question

webwork 5 - topics 10 - 12: problem 8
(1 point)
consider the integral
$$\int _ { 0 } ^ { 1 } \frac { 9 } { \sqrt { 1 - x ^ { 2 } } } d x$$
if the integral is divergent, type an upper - case \d\. otherwise, evaluate the integral.

Explanation:

Step1: Recall the integral formula

The integral of \(\frac{1}{\sqrt{1 - x^{2}}}\) is \(\arcsin(x)+C\). So, \(\int\frac{9}{\sqrt{1 - x^{2}}}dx=9\int\frac{1}{\sqrt{1 - x^{2}}}dx\).

Step2: Apply the fundamental theorem of calculus

By the fundamental theorem of calculus \(\int_{a}^{b}F^{\prime}(x)dx=F(b)-F(a)\). Here \(F(x) = 9\arcsin(x)\), \(a = 0\), and \(b = 1\). Then \(F(1)-F(0)=9\arcsin(1)-9\arcsin(0)\).
Since \(\arcsin(1)=\frac{\pi}{2}\) and \(\arcsin(0) = 0\).

Step3: Calculate the value

\(9\times\frac{\pi}{2}-9\times0=\frac{9\pi}{2}\)

Answer:

\(\frac{9\pi}{2}\)