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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 33

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Answer:  \frac{28}{3}

Hint: To solve this question we need to use  \int x^{n}  formula

Given:   \int_{1}^{3}\left|x^{2}-4\right| d x                        \left\{\begin{array}{l} x^{2}-4, x^{2} \geq 4 \\\\ -\left(x^{2}-4\right), x^{2} \leq 4 \end{array}\right.

\begin{aligned} &I=-\int_{1}^{2}\left(x^{2}-4\right) d x+\int_{2}^{3}\left(x^{2}-4\right) d x \\ & \end{aligned}

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

\begin{aligned} &I=-\left(\frac{8}{3}-8-\frac{1}{3}+4\right)+\left(\frac{27}{3}-12-\frac{8}{3}+8\right) \\ & \end{aligned}

I=-\frac{8}{3}+8+\frac{1}{3}-4+\frac{27}{3}-12-\frac{8}{3}+8 \\


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