IIT-JEE 1981Question 1
Find the area bounded by the curve $x^{2}=4y$ and the straight line $x=4y-2$.
Application of Integrals
Answer$\dfrac98$ sq. units
Solution
$y=\frac{x+2}4$ meets $y=\frac{x^2}4$ where $x^2-x-2=0$, i.e. $x=-1,2$. Area $=\frac14\int_{-1}^{2}(x+2-x^2)\,dx=\frac14\cdot\frac{27}6=\frac98$.