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2 ∫ 0 X √ 2 − X D X

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प्रश्न

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

\[Let\ I = \int_0^2 x\sqrt{2 - x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{2 - 2 + x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{x} d x\]
\[ = \int_0^2 \left( 2\sqrt{x} - x\sqrt{x} \right) dx\]
\[ = \int_0^2 \left( 2 x^\frac{1}{2} - x^\frac{3}{2} \right) dx\]
\[ = \left[ 2\frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^\frac{5}{2}}{\frac{5}{2}} \right]_0^2 \]
\[ = \left[ \frac{4}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} \right]_0^2 \]
\[ = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \]

`=(5xx8sqrt2)/(3xx5)-(3xx8sqrt2)/(5xx3)`

`=(16sqrt2)/15`

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अध्याय 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.5 | Q 33 | पृष्ठ ९५

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