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KEAM 2012 Math Paper

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Question : 14 of 120

Marks: +1, -0

The area bounded by

y = x + 2, y = 2 - x

and the x-axis is (in square units)

1
2
4
6
8

Solution:

Given lines are

y=x+2, \\;\\; y=2-x

The intersection point is

2-x=x+2 \Rightarrow 2 x=0

\Rightarrow \\;\\; x=0 \Rightarrow y=2

\therefore

Intersection point is

(0,2)

\text{therefore}

Required shaded region

=

Area of shaded region

A O B+

Area of shaded region

B O C

=\int\limits_{-2}^{0} y\\,dx+\int\limits_{0}^{2} y\\,dx

=\int\limits_{-2}^{0}(x+2)\\,dx+\int\limits_{0}^{2}(2-x)\\,dx

=\left[\\;\frac{x^2}{2}+2 x\right]_{-2}^{0}+\left[2 x-\\;\frac{x^2}{2}\right]^{2}_{0}

=\left[0+0-\left(\\;\frac{4}{2}-4\right)\right]+\left[4-\\;\frac{4}{2}\right]

=2+2=4\\,\text{sq}

units

Alternate method

\therefore

Area of bounded region

=

Area of

\triangle A B C

=\\;\frac{1}{2} \times A C \times O B

=\\;\frac{1}{2} \times 4 \times 2=4\\,\text{sq}

units

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