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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 1 Maths Textbook Solution.

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Answer: \frac{128}{15}

Given:  \int_{0}^{4} x \sqrt{4-x} d x

Hint: Use the formula  \int_{0}^{a} f(x) d x


\begin{aligned} &I=\int_{0}^{4} x \sqrt{4-x} d x \\ & \end{aligned}

I=\int_{0}^{4}(4-x) \sqrt{4-(4-x)} d x

\left(\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)

\begin{aligned} &=\int_{0}^{4}(4-x) \sqrt{x} d x \\ & \end{aligned}

=\int_{0}^{4} 4 \sqrt{x}-x^{\frac{3}{2}}

\begin{aligned} &=4\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{4}-\left(\frac{-x^{\frac{5}{2}}}{\frac{5}{2}}\right)_{0}^{4} \\ & \end{aligned}


\begin{aligned} &=\frac{64}{3}-\frac{64}{5} \\ & \end{aligned}

=\frac{320-192}{15} \\


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