Question

Find \(f\)

\(f'\left( x \right) = \frac{4}{{\sqrt {1 - {x^2}} }},\;f\left( {\frac{1}{2}} \right) = 1\)

Short answer

Required function is \(f\left( x \right) = 4{\sin ^{ - 1}}x + 1 - \frac{{2\pi }}{3}\).

Step-by-step solution

Find a general antiderivative of \(f'\)

Here we have,

\(f'\left( x \right) = \frac{4}{{\sqrt {1 - {x^2}} }}\)

Now, the general antiderivative of \(f'\left( x \right) = \frac{4}{{\sqrt {1 - {x^2}} }}\) is,

\(f\left( x \right) = 4{\sin ^{ - 1}}x + C\)

Find the value of constant \(C\)

Now, we have,\(f\left( x \right) = 4{\sin ^{ - 1}}x + C\)

Also, we have \(f\left( {\frac{1}{2}} \right) = 1\)

Now, by putting \(f\left( {\frac{1}{2}} \right) = 1\) in \(f\left( x \right) = 4{\sin ^{ - 1}}x + C\) we get,

\(\begin{aligned}{c}1 &= 4{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + C\\1 &= 4\left( {\frac{\pi }{6}} \right) + C\\C &= 1 - \frac{{2\pi }}{3}\end{aligned}\)

So, the required function is \(f\left( x \right) = 4{\sin ^{ - 1}}x + 1 - \frac{{2\pi }}{3}\).