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प्रश्न
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
योग
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उत्तर
\[\int\left( 1 - x \right)\sqrt{x} dx = \int\left( \sqrt{x} - x\sqrt{x} \right) dx\]
\[ = \int\left( x^\frac{1}{2} - x^\frac{3}{2} \right) dx\]
\[ = \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{x^\frac{3}{2} + 1}{\frac{3}{2} + 1} + c\]
\[ = \frac{2}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} + c\]
\[\text{ Hence,} \int\left( 1 - x \right)\sqrt{x} \text{ dx }= \frac{2}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} + c\]
\[ = \int\left( x^\frac{1}{2} - x^\frac{3}{2} \right) dx\]
\[ = \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{x^\frac{3}{2} + 1}{\frac{3}{2} + 1} + c\]
\[ = \frac{2}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} + c\]
\[\text{ Hence,} \int\left( 1 - x \right)\sqrt{x} \text{ dx }= \frac{2}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} + c\]
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