Question

Find f given the following information. f'(x)=4/sqrt(1-x^2) f(1/2)=1

Answer 1

Use the fundamental theorem of calculus (integrate).

Answer 2

This will give you \(f\) plus some constant. Then solve given the initial conditions given.

Answer 3

\[
f(b)-f(a) = \int _a^bf'(t)\;dt
\]In this case, we will let \(b = x\) and \(a=1/2\).\[
f(x)-f(1/2) = \int_{1/2}^xf'(t)\;dt= \int_{1/2}^x\frac{4}{\sqrt{1-t^2}}\;dt
\]

Answer 4

$$f(x) = \int \frac{4}{\sqrt{1-x^2}}\ dx = 4 \cdot \arcsin(x) + c$$
Solve for \(c\): \(f(\frac{1}{2}) = 4 \cdot \arcsin(\frac{1}{2}) + c = 1\)