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  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 30 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 30 maths textbook solution

Answers (1)

best_answer

Answer:

4

Hint:

To solve this we will split 1-x^{2} in differential form.

Given:

\int_{-2}^{2}\left|1-x^{2}\right| d x

Solution:

\begin{aligned} &\int_{-2}^{2}\left|1-x^{2}\right| d x \\\\ &1-x^{2}=0 \\\\ &x^{2}=1 \\\\ &x=\pm 1 \end{aligned}

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

\begin{aligned} &=\left[-\frac{1}{3}+1-\left(-\frac{8}{3}\right)+2\right]+\left[1-\frac{1}{3}-(-1)+\frac{1}{3}\right]+\left[\frac{8}{3}-2-\left(\frac{1}{3}-1\right)\right] \\\\ &=\frac{2}{3}-\left(-\frac{2}{3}\right)+\left[\frac{2}{3}-\left(-\frac{2}{3}\right)\right]+\left[\frac{2}{3}+\frac{2}{3}\right] \end{aligned}

\begin{aligned} &=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{12}{3} \\\\ &=4 \end{aligned}

 

 

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