Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 23

Answers (1)

Answer:  \frac{57}{5}-\sqrt{5}

Given:   \int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x

Hint: Use substitution method

Solution:   \int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x


\begin{aligned} &2 x+1=t^{2} \\ & \end{aligned}

 2 d x=2 t d t \\                   (Differentiate w.r.t to x)

d x=t d t

Now,  \int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x

\begin{aligned} &=\int_{2}^{4} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\left(\frac{t^{2}-1}{2}\right)}{t} t d t \\ & \end{aligned}

=\int_{2}^{4} \frac{\left(t^{2}-1\right)^{2}+2\left(t^{2}-1\right)}{4} d t

\begin{aligned} &=\frac{1}{4} \int_{2}^{4}\left(t^{4}+1-2 t^{2}+2 t^{2}-2\right) d t \\ & \end{aligned}

=\frac{1}{4} \int_{2}^{4} t^{4}-1 d t \\


\begin{aligned} &=\frac{1}{4}\left[\frac{(2 x+1)^{2} \sqrt{2 x+1}}{5}-\sqrt{2 x+1}\right]_{2}^{4} \\ & \end{aligned}

=\frac{1}{4}\left[\left(\frac{81 \times 3}{5}-3\right)-\left(\frac{25 \sqrt{5}}{5}-\sqrt{5}\right)\right]

\begin{aligned} &=\frac{1}{4}\left[\left(\frac{243-15}{5}\right)-\left(\frac{25 \sqrt{5}-5 \sqrt{5}}{5}\right)\right] \\ & \end{aligned}

=\frac{1}{4} \times \frac{228}{5}-\frac{1}{4} \times \frac{20 \sqrt{5}}{5} \\


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support