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

Answers (1)

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


               \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}



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