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Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 6

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Answer:  1

Hint: Use distribution method to find the value of x.

Given:   \int_{0}^{2}\left|x^{2}-3 x+2\right| d x


                \int_{0}^{2}\left|x^{2}-3 x+2\right| d x                                                            \left[\left(x^{2}+3 x+2\right)=(x-2)(x-1)\right]

Also we know that:

                |x|=\left\{\begin{array}{l} x, x \geq 0 \\ -x, x \leq 0 \end{array}\right\}

                \int_{0}^{1}\left(x^{2}-3 x+2\right) d x-\int_{1}^{2}\left(x^{2}-3 x+2\right) d x

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

                =\frac{1}{3}-\frac{3}{2}+2-\left[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\right] \\

                =\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2} \\


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