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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 9

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 9

Answers (1)

Answer:  5

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

Given:

                \begin{aligned} &\int_{-2}^{2}|x+1| d x \\ \end{aligned}

                x+1=0 \\

                x=-1

Solution:

                \begin{aligned} &\int_{-2}^{2}|x+1| d x \\ \end{aligned}

               \left\{\begin{array}{c} -(x+1),-2 \leq x \leq-1 \\ x+1, \quad-1 \leq x \leq 2 \end{array}\right\}

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

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

               \begin{aligned} &=\frac{-1}{2}+1+2-2+2+2-\frac{1}{2}+1 \\ & \end{aligned}

               =5

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