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  • Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 18

Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 18

Answers (1)

Answer:

9\; s q \cdot \text { units }

Hint:

Break the limit and find integral

Given:

y=\left | x+3 \right | and evaluate \int_{-6}^{0}|x+3| d x  What represent of graph

Solution:

Let's draw graph, y=\left | x+3 \right |

        y=|x+3|=\left\{\begin{array}{cc} x+3 & \text { for } x+3 \geq 0 \\ -(x+3) & \text { for } x+3<0 \end{array}\right.

        \left\{\begin{array}{c} x+3 \quad \text { for } x \geq-3 \\ -(x+3) \text { for } x+3<-3 \end{array}\right.

                                                    \left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

Required area  

        \begin{aligned} &=\int_{-6}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}|x+3| d x-=\int_{-3}^{0}|x+3| d x \\\\ &=\int_{-6}^{-3}-(x+3) d x-\int_{-3}^{0}(x+3) d x \end{aligned}

        \begin{aligned} &=\left[\frac{-x^{2}}{2}-3 x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3 x\right]_{-3}^{0} \\\\ &=\left[\frac{-(-3)^{2}}{2}-3(-3)\right]_{-6}^{-3}-\left[\frac{-(-6)^{2}}{2}-3(-6)\right]+\left[\frac{0^{2}}{2}+3(0)\right]-\left[\frac{(-3)^{2}}{2}+3(-3)\right] \end{aligned}

        \begin{aligned} &=\frac{-9}{2}-(-9)-\frac{-36}{2}-(-18)+\left[0+\left(\frac{-9}{2}+9\right)\right] \\\\ &=\frac{-9}{2}+9+0-\frac{9}{2}+9 \\\\ &=-9+18 \\\\ &=9 \text { sq.units } \end{aligned}

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