\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

\[\int_0^2 \sqrt{4 - x^2} d x\]\[ = \int_0^2 \sqrt{2^2 - x^2} d x\]\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{1}{2} \times 2^2 \sin^{- 1} \frac{x}{2} \right]_0^2 \]\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} \right]_0^2 + 2 \left[ \sin^{- 1} \frac{x}{2} \right]_0^2 \]\[ = 0 + 2\left( \frac{\pi}{2} - 0 \right)\]\[ = \pi\]

2 ∫ 0 √ 4 − X 2 D X - Mathematics

प्रश्न

\int_{0}^{2} \sqrt{4 - x^{2}} d x

उत्तर

$\int_{0}^{2} \sqrt{4 - x^{2}} d x$
$= \int_{0}^{2} \sqrt{2^{2} - x^{2}} d x$
$= {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{1}{2} \times 2^{2} \sin^{- 1} \frac{x}{2}]}_{0}^{2}$
$= {[\frac{x}{2} \sqrt{4 - x^{2}}]}_{0}^{2} + 2 {[\sin^{- 1} \frac{x}{2}]}_{0}^{2}$
$= 0 + 2 (\frac{π}{2} - 0)$
$= π$

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Definite Integrals

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Q 24 Q 23 Q 25

पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

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पाठ 20 Definite Integrals
Very Short Answers | Q 24 | पृष्ठ ११५

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2 ∫ 0 √ 4 − X 2 D X - Mathematics

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