Hi folks,

I am trying to evaluate the following integral $ \int_{0.294}^{0.956}\sqrt{4x - x^2} dx$

I checked the result in wolfram and get 0.94493

I have tried it myself and consistently get the same wrong answer. Can you spot the error?

$\int \sqrt{4 - (2 - x)^2} dx$

let $u = 2 - x$, so $\frac{du}{dx} = -1$ so $\int \sqrt{2^2 - u^2}. (-1) du$

let $u = 2 sin w$ , then $\frac{du}{dw} = 2 \cos w$ and $\sqrt{4 - u^2} = 2 cos w$

so that $ -4 \int \cos^2 w dw$

now $\cos^2 w = 1/2 (\cos 2w + 1)$

so

$-2 \int (\cos 2w + 1) dw = -2 ( 1/2 \sin 2w) - 2w = -\sin w - 2w = - u/2 - 2 \sin^{-1} (\frac{u}{2} )$

$ \frac{x - 2}{2} - 2 \sin^{-1} \frac{2 - x}{2} $

Now put in the limits

$ -0.522 - 2 \sin^{-1} (0.522) - (-0.853 - 2 \sin^{-1} (0.853) )$

$ -0.522 - 1.0984 + 0.853 + 2.0434 = -0.5631$

as you see a different result.