The area (in sq.

Question

The area (in sq. units) of the region

{x $$ \in $$ R : x $$ \ge $$ 0,  y $$ \ge $$ 0, y $$ \ge $$ x - 2  and y $$ \le $$ $$\sqrt x $$}, is :

Accepted Answer

y = $$\sqrt x $$

y = x - 2

$$\therefore\,\,\,$$ $$\sqrt x $$ = x - 2

$$ \Rightarrow $$ $\,\,\,$x = x2$-$4x + 4

x2$-$5x + 4 = 0

x2$-$4x$-$x + 4 = 0$ \Rightarrow $$ $\,\,\,$$ x(x - 4) - (x - 4) = 0

$$ \Rightarrow $$ $\,\,\,$(x$-$4) (x$-$1) = 0$\therefore\,\,\,$x = 4, 1

and y = 2,$-$1$\therefore\,\,\,$$ Their point of intersection (4, 2) and (1, $$-$$ 1)

Required area is shown in the shaded figure.

$$\therefore\,\,\,$Required area

=$\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x  - x + 2} \right)\,dx} } $=$\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} } $=$\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$=${2 \over 3}\left( 8 \right) ( + 2\]\left( {4 - 2} \right) ( \]-$${1 \over 2}$(16$-$4)

=${{16} \over 3}$+ 4$-$6

=${{10} \over 3}$

Solutions · 1