#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 18

Answer:  $\frac{63}{2}$

Hint: You must know the rules of solving definite integral.

Given:  $\int_{-5}^{0} f(x) d x \text { where } f(x)=|x|+|x+2|+|x+5|$

Solution:

For the first integrand:

\begin{aligned} &x<0 \\ & \end{aligned}

$|x|=-x$

\begin{aligned} \int_{-5}^{0}|x| d x & \\ & \end{aligned}

$=\int_{-5}^{0}-x d x \\$

$=-\left[\frac{x^{2}}{2}\right]_{-5}^{0} \\$

$=\frac{25}{2}$

For the second integrand:

$\begin{gathered} (x+2)=\left\{\begin{array}{l} x+2, \text { where } x \geq-2 \\\\ -(x+2), \text { wherex } \leq-2 \end{array}\right\} \\ \end{gathered}$

$\int_{-5}^{0}|x+2| d x=\int_{-5}^{-2}-(x+2) d x+\int_{-2}^{0}(x+2) d x$

For the third integrand:

$\begin{array}{r} |x+5|=x+5, \text { if } x \geq-5 \\ \end{array}$

$\int_{-5}^{0}|x+5| d x=\int_{-5}^{0}(x+5) d x$

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

$=0-\frac{25}{2}+25 \\$

$=\frac{25}{2}$

Hence the total integration will be

\begin{aligned} \int_{-5}^{0} f(x) d x &=\int_{-5}^{0}|x| d x+\int_{-5}^{0}|x+2| d x+\int_{-5}^{0}|x+5| d x \\ & \end{aligned}

$=\frac{25}{2}+\frac{13}{2}+\frac{25}{2} \\\\$

$=\frac{63}{2}$