Question

integral of sqrt(4x-x^2)dx

Answer 1

Complete the square:
\[4x-x^2=-\left(x^2-4x\right)=-\left(x^2-4x+4-4\right)=4-\left(x-2\right)^2\]
\[\int\sqrt{4-(x-2)^2}~dx\]
You've learned trig subs, correct?

Answer 2

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
Let \(x-2=2\sin u\), so that \(dx=2\cos u~du\):
\[\int\sqrt{4-\left(2\sin u\right)^2}~(2\cos u~du)\\
\int\sqrt4\sqrt{1-\sin^2 u}~(2\cos u~du)\\
4\int\sqrt{1-\sin^2 u}~\cos u~du\]
From here, you can simplify.
\(\color{blue}{\text{End of Quote}}\)

Answer 3

\[\int\limits\limits(\sqrt{4-(x-2)^2} )dx \]


\[(x-2)/2*\sqrt{4-(x-2)^2} + 2\sin^{-1} ((x-2)/4) +C\]

Answer 4

Standard Formula