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

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

Answers (1)

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 \\

I=\frac{28}{3}

Posted by

infoexpert27

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