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KEAM 2010 Math Paper

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Question : 50 of 120

Marks: +1, -0

The area of the plane region bounded by the curve

x=y^{2}-2

y=-x

is (in square units)

$\; \frac{13}{3}$
$\; \frac{2}{5}$
$\; \frac{9}{2}$
$\; \frac{5}{2}$
$\; \frac{13}{2}$

Solution:

Given curves

x=y^{2}-2

and

y=x

Thus, interection point are

(-1,1)

and

\;\;(2,-2)

We are to find the area of shaded part

Area of

\text{A B C} = \int\limits_{-2}^{-1} \sqrt{x+2} d x

=\left[ \frac{2}{3} (x+2)^{3/2} \right]_{-2}^{-1} = \frac{2}{3} \text{ sq}

unit

Area of

\text{B C O} = \int\limits_{-1}^{0} -x d x = \left( -\frac{x^{2}}{2} \right)_{-1}^{0}

= \; \frac{1}{2} \text{ sq}

unit

Area of ADO

= \int\limits_{-2}^{0} \sqrt{x+2} d x = \left[ \frac{2}{3} (x+2)^{3/2} \right]_{-2}^{0}

= \; \frac{4}{3} \sqrt{2}

Area of

\text{ODE} =

area of

\text{ODEF-}

area of

\text{OFE}

\int\limits_{0}^{2} \sqrt{x+2} d x - \int\limits_{0}^{2} (-x) d x

= \left\{ \frac{2}{3} (x+2)^{3/2} \right\}_{0}^{2} - \left( -\frac{x^{2}}{2} \right)_{0}^{2}

= \left( \frac{16}{3} - \frac{4 \sqrt{2}}{3} \right) - (2)

(neglecting the negative sign)

= \; \frac{16}{3} - \frac{4 \sqrt{2}}{3} - 2 \text{ sq}

unit

\therefore

Required area

= \; \frac{2}{3} + \; \frac{1}{2} + \; \frac{4 \sqrt{2}}{3} + \; \frac{16}{3} - \; \frac{4 \sqrt{2}}{3} - 2

= \; \frac{2}{3} + \; \frac{1}{2} + \; \frac{16}{3} - 2

= \; \frac{27}{6} = \; \frac{9}{2} \text{ sq}

unit

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