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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 2 Maths Textbook Solution.

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Answer: \frac{326}{135}

Given:  \int_{1}^{2} x \sqrt{3 x-2} d x

Hint: Use the substitution method

Solution: \int_{1}^{2} x \sqrt{3 x-2} d x

Let  3 x-2=t

3dx = dt (differentiating w.r.t  x)

\begin{aligned} &\Rightarrow \int_{1}^{2} \frac{t+2}{3} \times \sqrt{t} \times \frac{d t}{3} \\ & \end{aligned}

=\frac{1}{9} \int_{1}^{2}(t+2) \sqrt{t} d t \\

=\frac{1}{9} \int_{1}^{2}(t)^{\frac{3}{2}}+2 \sqrt{t} d t

\begin{aligned} &=\frac{1}{9}\left[\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)_{1}^{2}+2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)_{1}^{2}\right] \\ & \end{aligned}

=\frac{1}{9}\left\{\frac{2}{5}\left[(3 x-2)^{\frac{5}{2}}\right]_{1}^{2}+\frac{4}{3}\left[(3 x-2)^{\frac{3}{2}}\right]_{1}^{2}\right\}

\begin{aligned} &=\frac{1}{9}\left[\frac{2}{5}(32-1)+\frac{4}{3}(8-1)\right] \\ & \end{aligned}

=\frac{1}{9}\left(\frac{2}{5} \times 31+\frac{4}{3} \times 7\right)=\frac{1}{9}\left(\frac{186+140}{15}\right) \\




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