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प्रश्न
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
उत्तर
\[\int_1^2 x\sqrt{3x - 2} d x\]
\[Let, 3x - 2 = t,\text{ then }3dx = dt\]
\[\text{when, }x = 1 ; t = 1\text{ and }x = 2 ; t = 4\]
\[\text{Therefore the integral becomes}\]
\[ \int_1^4 \frac{t + 2}{3}\sqrt{t} \frac{dt}{3}\]
\[ = \frac{1}{9} \int_1^4 t^\frac{3}{2} + 2\sqrt{t} dt\]
\[ = \frac{1}{9} \left[ \frac{2 t^\frac{5}{2}}{5} + \frac{4 t^\frac{3}{2}}{3} \right]_1^4 \]
\[ = \frac{1}{9}\left[ \frac{64}{5} + \frac{32}{3} - \frac{2}{5} - \frac{4}{3} \right]\]
\[ = \frac{46}{135}\]
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