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

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

Answers (1)

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}

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