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#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 3

Answer:  $10$

Hint: Break the range of integration like this   $\int_{-3}^{-1} f(x) \& \int_{-1}^{3} f(x)$

Given:  $\int_{-3}^{3}|x+1| d x$

Solution:

\begin{aligned} &I=\int_{-3}^{3}|x+1| d x \\ & \end{aligned}

$f(x)=|x+1|=\left\{\begin{array}{ll} -(x+1), & \text { if }-3 \leq x \leq-1 \\ (x+1), & \text { if }-1 \leq x \leq 3 \end{array}\right\}$

\begin{aligned} &I=\int_{-3}^{-1} f(x) d x+\int_{-1}^{3} f(x) d x \\ & \end{aligned}

$I=\int_{-3}^{-1}-(x+1) d x+\int_{-1}^{3}(x+1) d x$

\begin{aligned} &I=\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{-3}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{3} \\ & \end{aligned}                               $\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

$y=-\left[\frac{(1-9)}{2}+(-1+3)\right]+\frac{(9-1)}{2}+3+1$

\begin{aligned} &=-(-4+2)+(4+4) \\ & \end{aligned}

$=-(-2)+8 \\$

$=2+8 \\$

$=10$