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4 ∫ 0 1 √ 16 − X 2 D X . - Mathematics

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प्रश्न

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]
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उत्तर

\[\int_0^4 \frac{1}{\sqrt{16 - x^2}} d x\]

\[ = \int_0^4 \frac{1}{\sqrt{4^2 - x^2}} d x\]

\[ = \left[ \sin^{- 1} \frac{x}{4} \right]_0^4 \]

\[ = \left( \frac{\pi}{2} - 0 \right)\]

\[ = \frac{\pi}{2}\]

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

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Very Short Answers | Q 11 | पृष्ठ ११५

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