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  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 2

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 2

Answers (1)

Answer:  20

Hint: Break the range of integration like this   \int_{-4}^{-2} f(x) \& \int_{-2}^{4} f(x)

Given:

                I=\int_{-4}^{4}|x+2| d x

Solution:

                \begin{aligned} &f(x)=\left\{\begin{array}{c} -v e, \quad \text { if }-4 \leq x \leq-2 \\ +v e, \text { if }-2 \leq x \leq 4 \end{array}\right\} \\ & \end{aligned}

               I=\int_{-4}^{-2} f(x) d x+\int_{-2}^{4} f(x) d x

                  \begin{aligned} &=\int_{-4}^{-2}-(x+2) d x+\int_{-2}^{4}(x+2) d x \\ & \end{aligned}

                 =-\left[\frac{x^{2}}{2}+2 x\right]_{-4}^{-2}+\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{4}                                                \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

                 \begin{aligned} &=-\left(\frac{1}{2}(4-16)+2(-2+4)\right)+\frac{1}{2}(16-4)+2(4+2) \\ & \end{aligned}

                =-(-6+4)+6+12 \\

                =2+18 \\

                =20

 

 

           

Posted by

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