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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 19

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 19

Answers (1)

Answer:

9\; s q \cdot { units }

Hint:

Find integral \int_{-4}^{2}|x+1| d x

Given:

y=|x+1|  Evaluate \int_{-4}^{2}|x+1| what does the value of integral represented on graph

Solution:

We have,

       y=|x+1| intersect x = -4 and x = 2 at ( -4,3 ) and ( 2,3 ).

Now,

y=|x+1|

    =\left\{\begin{array}{c} (x+1) \text { for all } x>-1 \\ -(x+1) \text { for all } x<-1 \end{array}\right.

Integral represents the area enclosed between x = -4 and x = 2.

\begin{aligned} A &=\int_{-4}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}|y| d x+\int_{-1}^{2}|y| d x \\\\ A &=\int_{-4}^{-1}-(x+1) d x+\int_{-1}^{2}(x+1) \end{aligned}

\begin{aligned} &A=-\left[\frac{x^{2}}{2}+x\right]_{-4}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2} \\\\ &A=-\left[\frac{1}{2}-1-\frac{16}{2}+4\right]+\left[\frac{4}{2}+2-\frac{1}{2}+1\right] \end{aligned}

\begin{aligned} &A=-\left[3-\frac{15}{2}\right]+\left[5-\frac{1}{2}\right] \\\\ &A=-3+\frac{15}{2}+5-\frac{1}{2} \\\\ &A=\frac{-6+15+10-1}{2} \end{aligned}

A=9\; s q \cdot { units }

Posted by

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