JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 22)
The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :
$${3 \over 4}$$
$${5 \over 4}$$
$${7 \over 8}$$
$${9 \over 8}$$
Explanation
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x = 4y $$-$$ 2 & x2 = 4y
$$ \Rightarrow $$ x2 = x + 2 $$ \Rightarrow $$ x2 $$-$$ x $$-$$ 2 = 0
x = 2, $$-$$ 1
So, $$\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}} $$
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