Evaluate :$\int_{-1}^{1} 2\sqrt{1-x^2} dx $

Geometrically , the unit circle can be represented as $$x^2+y^2=1$$

so $$y=\pm \sqrt{1-x^2}$$ and your case $y=+ \sqrt{1-x^2}$

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So $\int_{-1}^1\sqrt{1-x^2} dx $ is the area of a (upper )semi circle, which is $\frac{\pi}{2}$. So $$2 \int_{-1}^1 \sqrt{1-x^2}dx =2 \frac{\pi}{2}=\pi$$


Before evaluating the definite integral, you need to make the following back-substitution: $$ u=\arcsin{x} $$

And use this trigonometric identity ($-\frac{\pi}{2}\le u\le \frac{\pi}{2} \implies \cos{u}\ge0$):

$$ \sin{2u}=2\sin{u}\cos{u}=2\sin{(\arcsin{x})}\sqrt{1-\sin^2{(\arcsin{x})}}=2x\sqrt{1-x^2} $$

You also forgot that you have a $2$ in front of your integral:

$$ 2\left[\frac{\arcsin{x}}{2}+x\sqrt{1-x^2}\right]_{-1}^{1}=\\ 2\left(\frac{\arcsin{(1)}}{2}+1\cdot\sqrt{1-1^2}-\frac{\arcsin{(-1)}}{2}-1\cdot\sqrt{1-(-1)^2}\right)=\\ 2\left(\frac{\pi}{4}+0+\frac{\pi}{4}-0\right)=2\frac{2\pi}{4}=\frac{4\pi}{4}=\pi. $$

Tags:

Integration

Calculus

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