JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 16)
The area (in sq. units) of the region
A = {(x, y) : x2 $$ \le $$ y $$ \le $$ x + 2} is
A = {(x, y) : x2 $$ \le $$ y $$ \le $$ x + 2} is
$${{31 \over 6}}$$
$${{10 \over 3}}$$
$${{13 \over 6}}$$
$${{9 \over 2}}$$
Explanation
Parabola : x2 = y
Straight line : y = x + 2
$$ \therefore $$ x2 = x + 2
$$ \Rightarrow $$ x2 - x - 2 = 0
x = -1, 2
$$ \therefore $$ y = 1, 4_9th_April_Morning_Slot_en_16_2.png)
Required area = $$\int\limits_{ - 1}^2 {\left[ {\left( {x + 2} \right) - {x^2}} \right]} dx$$
= $$\left[ {{{{x^2}} \over 2} + 2x - {{{x^3}} \over 3}} \right]_{ - 1}^2$$
= $$\left( {2 + 4 - {8 \over 3}} \right) - \left( {{1 \over 2} - 2 + {1 \over 3}} \right)$$
= $${{9 \over 2}}$$
Straight line : y = x + 2
$$ \therefore $$ x2 = x + 2
$$ \Rightarrow $$ x2 - x - 2 = 0
x = -1, 2
$$ \therefore $$ y = 1, 4
Required area = $$\int\limits_{ - 1}^2 {\left[ {\left( {x + 2} \right) - {x^2}} \right]} dx$$
= $$\left[ {{{{x^2}} \over 2} + 2x - {{{x^3}} \over 3}} \right]_{ - 1}^2$$
= $$\left( {2 + 4 - {8 \over 3}} \right) - \left( {{1 \over 2} - 2 + {1 \over 3}} \right)$$
= $${{9 \over 2}}$$
Comments (0)
