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  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 20

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 20

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Answer:  \frac{19}{2}

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

Given:  \int_{-1}^{2}|x+1|+|x|+|x-1| d x

Solution:

               \begin{aligned} &f(x)=|x+1|+|x|+|x-1| \\ & \end{aligned}

              f(x)=x+1-x-x+1=-x+2 \\

              f(x)=x+1+x-x+1 \rightarrow 0 \leq x \leq 1=2+x

              \begin{aligned} &f(x)=x+1+x+x-1 \rightarrow x \geq 1=3 x \\ & \end{aligned}

              f(x)=-x-1-x-x+13-3 x=0

              \begin{aligned} &=\int_{-1}^{0}(2-x) d x+\int_{0}^{1}(x+2) d x+\int_{1}^{2} 3 x d x \\ & \end{aligned}

              =\left[2 x-\frac{x^{2}}{2}\right]_{-1}^{0}+\left(\frac{x^{2}}{2}+2 x\right)_{0}^{1}+\left(\frac{3 x^{2}}{2}\right)_{1}^{2}

              \begin{aligned} &=2+\frac{1}{2}+\frac{1}{2}+2+6-\frac{3}{2} \\ & \end{aligned}

              =\frac{19}{2}

        

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