#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 4

Answer:  $\frac{5}{2}$

Hint: Break the range of integration and then integrate.

Given:  $\int_{-1}^{1}|2 x+1| d x$

Solution:

$f(x)=2 x+1=\left\{\begin{array}{ll} -(2 x+1), & \text { if }-1 \leq x \leq \frac{-1}{2} \\ (2 x+1), & \text { if } \frac{-1}{2} \leq x \leq 1 \end{array}\right\}$

\begin{aligned} &I=\int_{-1}^{\frac{-1}{2}} f(x) d x+\int_{\frac{-1}{2}}^{1} f(x) d x \\ & \end{aligned}

$I=\int_{-1}^{\frac{-1}{2}}-(2 x+1) d x+\int_{\frac{-1}{2}}^{1}(2 x+1) d x$

Using the formula: $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

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

$I=-\left[\left(x^{2}+x\right)\right]_{-1}^{\frac{-1}{2}}+\left[x^{2}+x\right]_{\frac{-1}{2}}^{1}$

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

$=\frac{-1}{4}+\frac{1}{2}+2-\frac{1}{4}+\frac{1}{2} \\$

$=\frac{-2}{4}+\frac{2}{2}+2 \\$

$=\frac{5}{2}$