JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 8)
The area (in sq. units) of the region
A = {(x, y) : (x – 1)[x] $$ \le $$ y $$ \le $$ 2$$\sqrt x $$, 0 $$ \le $$ x $$ \le $$ 2}, where [t]
denotes the greatest integer function, is :
A = {(x, y) : (x – 1)[x] $$ \le $$ y $$ \le $$ 2$$\sqrt x $$, 0 $$ \le $$ x $$ \le $$ 2}, where [t]
denotes the greatest integer function, is :
$${8 \over 3}\sqrt 2 - 1$$
$${4 \over 3}\sqrt 2 + 1$$
$${8 \over 3}\sqrt 2 - {1 \over 2}$$
$${4 \over 3}\sqrt 2 - {1 \over 2}$$
Explanation
y = (x – 1)[x] = $$\left\{ {\matrix{
{0,} & {0 \le x < 1} \cr
{x - 1,} & {1 \le x < 2} \cr
{2,} & {x = 2} \cr
} } \right.$$
_5th_September_Evening_Slot_en_8_1.png)
A = $$\int\limits_0^2 {2\sqrt x } dx - {1 \over 2}.1.1$$
= 2$$\left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2$$ - $${1 \over 2}$$
= $${{8\sqrt 2 } \over 3} - {1 \over 2}$$
A = $$\int\limits_0^2 {2\sqrt x } dx - {1 \over 2}.1.1$$
= 2$$\left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2$$ - $${1 \over 2}$$
= $${{8\sqrt 2 } \over 3} - {1 \over 2}$$
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