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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 7

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 7

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Answer:  \frac{13}{2}

Hint: Use   x=\frac{1}{2},

                \begin{aligned} &2 x-1=0 \\ & \end{aligned}

                x=\frac{1}{2}

Given:     \int_{0}^{3}|2 x-1| d x

                                I=\int_{0}^{3} f(x) d x 

                                f(x)=\left\{\begin{array}{ll} -(2 x-1), & \text { if } 0 \leq x \leq \frac{1}{2} \\ (2 x-1), & \text { if } \frac{1}{2} \leq x \leq 3 \end{array}\right\}
                                =\int_{0}^{\frac{1}{2}}-(2 x-1) d x+\int_{\frac{1}{2}}^{3}(2 x-1) d x
                                I=\left[-\left(x^{2}-x\right)\right]_{0}^{\frac{1}{2}}+\left(x^{2}-x\right)_{\frac{1}{2}}^{3}

                               \begin{aligned} I &=-\left[\left(\frac{1}{2}\right)^{2}-\frac{1}{2}-0\right]+\left[(3)^{2}-3-\left(\left(\frac{1}{2}\right)^{2}-\frac{1}{2}\right)\right] \\ \end{aligned}

                              =\frac{-1}{4}+\frac{1}{2}+6-\frac{1}{4}+\frac{1}{2} \\

                             =\frac{-1}{2}+1+6 \\

                             =\frac{-1+14}{2} \\

                              =\frac{13}{2}

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