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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise 19.3 Question 5

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Answer:   \frac{25}{2}

Hint: Break the range of integration and then solve the integration.

Given:  \int_{-2}^{2}|2 x+3| d x


                f(x)=|2 x+3|                

                f(x)=\left\{\begin{array}{ll} -(2 x+3), & \text { if }-2 \leq x \leq \frac{-3}{2} \\ (2 x+3), & \text { if } \frac{-3}{2} \leq x \leq 2 \end{array}\right\}                                                 \quad\left(\begin{array}{c} 2 x+3=0 \\ x=\frac{-3}{2} \end{array}\right)

                \begin{aligned} &I=\int_{-2}^{\frac{-3}{2}} f(x) d x+\int_{\frac{-3}{2}}^{2} f(x) d x \\ & \end{aligned}

                I=\int_{-2}^{\frac{-3}{2}}-(2 x+3) d x+\int_{\frac{-3}{2}}^{2}(2 x+3) d x

Use the formula: \left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

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

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

                 =-\left(\frac{9}{4}-4-\frac{9}{2}+6\right)+\left(4-\frac{9}{4}+6+\frac{9}{2}\right) \\

                = \frac{25}{2}



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