JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 17)
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 :
{x $$ \in $$ R : x $$ \ge $$ 0, y $$ \ge $$ 0, y $$ \ge $$ x $$-$$ 2 and y $$ \le $$ $$\sqrt x $$}, is :
$${{13} \over 3}$$
$${{8} \over 3}$$
$${{10} \over 3}$$
$${{5} \over 3}$$
Explanation
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)
_15th_April_Morning_Slot_en_17_1.png)
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
= $${{10} \over 3}$$
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
= $${{10} \over 3}$$
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