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  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 10

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 10

Answers (1)

Answer:

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

Given:     \begin{aligned} &\int_{1}^{2}|x-3| d x \\ & \end{aligned}

                x-3=0 \\

                x=3

Solution:

                \begin{aligned} \mathrm{I} &=\int_{1}^{2}|x-3| d x \\ & \end{aligned}

                =\int_{1}^{2}-(x-3) d x \\

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

                \begin{aligned} &l=-\left(\frac{4}{2}-6-\frac{1}{2}+3\right) \\ & \end{aligned}

                I=-\left(\frac{4}{2}-6-\frac{1}{2}+3\right) \\

                I=-\left(\frac{3}{2}-3\right)

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

                =\frac{-3+6}{2} \\

                 =\frac{3}{2}

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