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  • Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 2

Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 2

Answers (1)

Answer: \frac{17}{2}sq\cdot units

Hint: Use   - x+ y = 1

Given: Using integration, find area of region bounded by line y-1=x the x-axis and ordinatex=-2 and x=3

Solution:

Here y-1=x is equation of line

We can write

\begin{aligned} &y=x+1 \\\\ &-x+y=1 \end{aligned}

We get \frac{-x}{1}+\frac{y}{1}=1

Consider AB as line intersecting the x-axis at point C(-1,0)

So required area=\text { Area of } C D A C+\text { Area of } C B E C

=\int_{-1}^{3} y d x+\int_{-2}^{-1}-(y) d x

Substituting the value of y,

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

Substituting the value of x ,

\begin{aligned} &=\left(\frac{9}{2}+3\right)-\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}-1\right)-(2-2) \\\\ &=\left(\frac{15}{2}+\frac{1}{2}\right)-\left(\frac{-1}{2}\right) \\\\ &=8+\frac{1}{2} \\ &=\frac{17}{2} s q \cdot \text { units } \end{aligned}

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