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1 ∫ 0 √ X ( 1 − X ) D X - Mathematics

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Question

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\]
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Solution

\[Let\ I = \int_0^1 \sqrt{x\left( 1 - x \right)} d x . Then, \]
\[I = \int_0^1 \sqrt{\frac{1}{4} - \left( x - \frac{1}{2} \right)^2} dx\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \sqrt{1 - \frac{\left( x - \frac{1}{2} \right)^2}{\frac{1}{4}}} dx\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \sqrt{1 - \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right)^2} dx\]
\[Let \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right) = \sin u\]
\[ \Rightarrow 2 dx = \cos u du\]
\[ \therefore I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sqrt{1 - \sin^2 u} \cos u du\]
\[ \Rightarrow I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos^2 u du\]
\[ \Rightarrow I = \frac{1}{4} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( \frac{\cos 2u + 1}{2} \right) du\]
\[ \Rightarrow I = \frac{1}{8} \left[ \frac{\sin 2u}{2} + u \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{8}\left[ \frac{\pi}{2} + \frac{\pi}{2} \right]\]
\[ \Rightarrow I = \frac{\pi}{8}\]

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Definite Integrals
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Chapter 20: Definite Integrals - Exercise 20.1 [Page 17]

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RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.1 | Q 41 | Page 17

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