Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_{0}^{a}(a-x)^{\frac{3}{2}} \cdot x^{2} d x=$
Options:
Solution:
1297 Upvotes
Verified Answer
The correct answer is:
$\frac{16 a^{\frac{9}{2}}}{315}$
Let
$\begin{aligned} I &=\int_{0}^{a}(a-x)^{\frac{3}{2}} x^{2} d x \\ &=\int_{0}^{a}[a-(a-x)]^{\frac{3}{2}}(a-x)^{2} d x \\ &=\int_{0}^{a} x^{\frac{3}{2}}\left(a^{2}-2 a x+x^{2}\right) d x=\int_{0}^{a}\left(a^{2} x^{\frac{3}{2}}-2 a x^{\frac{5}{2}}+x^{\frac{7}{2}}\right)_{d x} \\ &=a^{2}\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{a}-2 a\left[\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{a}+\left[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}\right]_{0}^{a} \\ &=\frac{2 a^{2}}{5}\left[a^{\frac{5}{2}}-0\right]-2 a \times \frac{2}{7}\left[a^{\frac{7}{2}}-0\right]+\frac{2}{9}\left[a^{\frac{9}{2}}-0\right]=\frac{2}{5} a^{\frac{9}{2}}-\frac{4}{7} a^{\frac{9}{2}}+\frac{2}{9} a^{\frac{9}{2}} \\ &=a^{\frac{9}{2}}\left[\frac{2}{5}-\frac{4}{7}+\frac{2}{9}\right]=\frac{16}{315} a^{\frac{9}{2}} \end{aligned}$
$\begin{aligned} I &=\int_{0}^{a}(a-x)^{\frac{3}{2}} x^{2} d x \\ &=\int_{0}^{a}[a-(a-x)]^{\frac{3}{2}}(a-x)^{2} d x \\ &=\int_{0}^{a} x^{\frac{3}{2}}\left(a^{2}-2 a x+x^{2}\right) d x=\int_{0}^{a}\left(a^{2} x^{\frac{3}{2}}-2 a x^{\frac{5}{2}}+x^{\frac{7}{2}}\right)_{d x} \\ &=a^{2}\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{a}-2 a\left[\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{a}+\left[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}\right]_{0}^{a} \\ &=\frac{2 a^{2}}{5}\left[a^{\frac{5}{2}}-0\right]-2 a \times \frac{2}{7}\left[a^{\frac{7}{2}}-0\right]+\frac{2}{9}\left[a^{\frac{9}{2}}-0\right]=\frac{2}{5} a^{\frac{9}{2}}-\frac{4}{7} a^{\frac{9}{2}}+\frac{2}{9} a^{\frac{9}{2}} \\ &=a^{\frac{9}{2}}\left[\frac{2}{5}-\frac{4}{7}+\frac{2}{9}\right]=\frac{16}{315} a^{\frac{9}{2}} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.