# Find f. f'(x)= \frac{8}{\sqrt{1 - x2}},\ f(1/2)= 4

## Question:

Find f.

{eq}f'(x)= \frac{8}{\sqrt{1 - x2}},\ f(1/2)= 4 {/eq}

## Indefinite Integral:

Assume that {eq}v {/eq} is a single-valued function of {eq}x {/eq} and {eq}K {/eq} is an anti derivative of {eq}v {/eq} such that {eq}\displaystyle v\left( x \right) = K'\left( x \right),\;\forall \;x {/eq}.

Then the indefinite integral of {eq}\displaystyle v {/eq} is defined as {eq}\displaystyle \int v\left( x \right)dx = K\left( x \right) + C {/eq}, where {eq}c {/eq} is an integration constant.

## Answer and Explanation:

Given that: {eq}\displaystyle f'(x) = \frac{8}{{\sqrt {1 - {x^2}} }},{\text{ }}f(1/2) = 4 {/eq}

{eq}\displaystyle \eqalign{ & f'(x) = \frac{8}{{\sqrt {1 - {x^2}} }},{\text{ }}f(1/2) = 4 \cr & f'(x) = \frac{8}{{\sqrt {1 - {x^2}} }} \cr & {\text{Taking integration both side;}} \cr & \int {df(x)} = \int {\frac{8}{{\sqrt {1 - {x^2}} }}dx} \cr & f(x) = 8\int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} \cr & f(x) = 8{\sin ^{ - 1}}x + c \cr & f(x) = 8{\sin ^{ - 1}}x + c \cr & \cr & f\left( {\frac{1}{2}} \right) = 8{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + c \cr} {/eq}