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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 20 maths

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Answer: 2

Hint: You must know about \left | x \right | function.

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


Let I=\int_{-1}^{1}|1-x| d x

We know |1-x|=\left\{\begin{array}{c} 1-x \quad ; \quad 1-x \geq 0 \\ -x \geq-1 \Rightarrow x \leq 1 \\\\ -(1-x) ; \quad 1-x<0 \\ -x<-1 \Rightarrow x>1 \end{array}\right\}

\therefore \int_{-1}^{1}|1-x| d x=\int_{-1}^{1}(1-x) d x

                               \begin{aligned} &=\int_{-1}^{1} 1 d x-\int_{-1}^{1} x \; d x \\\\ &=\left[x-\frac{x^{2}}{2}\right]_{-1}^{1} \end{aligned}

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



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