Prove-1-1-1-x-2-dx-pi-2-

Question Number 77474 by 21042004 last updated on 06/Jan/20

P r o v e

\int_{- 1}^{1} \sqrt{1 - x^{2}} d x = \frac{π}{2}

Commented by mathmax by abdo last updated on 06/Jan/20

l e t I = \int_{- 1}^{1} \sqrt{1 - x^{2}} d x \Rightarrow I = 2 \int_{0}^{1} \sqrt{1 - x^{2}} d x c h a n g e m e n t x = s i n θ

g i v e I = 2 \int_{0}^{\frac{π}{2}} c o s θ c o s θ d θ = \int_{0}^{\frac{π}{2}} (1 + c o s (2 θ) d θ

= \frac{π}{2} + {[\frac{1}{2} s i n (2 θ)]}_{0}^{\frac{π}{2}} = \frac{π}{2} + 0 = \frac{π}{2} .

Answered by MJS last updated on 06/Jan/20

(1) y = \sqrt{1 - x^{2}} is the upper half of the circle

with center (\begin{matrix} 0 \\ 0 \end{matrix}) and radius r = 1 \Rightarrow the integral

is half of the circle area = \frac{1}{2} r^{2} π = \frac{π}{2}

(2) \int \sqrt{1 - x^{2}} d x =

[t = \arcsin x \to d x = \sqrt{1 - x^{2}} d t]

= \int \cos^{2} t d t = \frac{1}{2} (t + \sin t \cos t) =

= \frac{1}{2} (x \sqrt{1 - x^{2}} + \arcsin x) + C

\Rightarrow the integral is \frac{π}{2}

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