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  • Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.

Q: Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.

Answers (1)

The lines given are x + 2y = 2 

x = 2 - 2y \ldots .(1) \\

y – x = 1 

x = y -1 \ldots .(2) \\

and  2x + y = 7 \ldots .(3) \\

Equate the values of x from 1 and 2 to get,

\\$2-2 y=y-1$\\ $2+1=y+2 y$\\ $3=3 y$\\ $y=1$\\

put the value of y in (2) to get

\\$x=1-1$\\ =0\\

So, intersection point is (0,1)

On solving the equations, the point of interaction is found to be (0, 1), (2, 3) and (4, -1) 

Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by A=\int_{a}^{b} f(x) d x \: \: or\: \: \int_{a}^{b} y d x

Required area

$$ =\int_{0}^{2}\left(\mathrm{x}-1-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx}+\int_{2}^{4}\left(7-2 \mathrm{x}-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx} $$ \\
\\ =\int_{0}^{2}\left(\frac{3 x}{2}\right) d x+\int_{2}^{4}\left(6-\frac{3 x}{2}\right) d x \\ =\left[\frac{3 x^{2}}{4}\right]_{0}^{2}+\left[6 x-\frac{3 x^{2}}{4}\right]_{2}^{4} \\ =3+(24-12)-(12-3)=6 \text { sq.units }

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infoexpert22

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