Question

Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3, is
(a) 2

(b) $\frac{9}{4}$

(c) $\frac{9}{3}$

(d) $\frac{9}{2}$

Answer 1

$\left(\mathrm{b}\right)\frac{9}{4}$



y² = 4x represents a parabola with vertex at origin O(0, 0) and symmetric about +ve x-axis

y = 3 is a straight line parallel to the x-axis

Point of intersection of the line and the parabola is given by
Substituting y = 3 in the equation of the parabola
$$\begin{gathered} y^2=4x \\ \Rightarrow 3^2=4x \\ \Rightarrow x=\frac{9}{4} \\ \mathrm{Thus}A\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,3\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{and}\mathrm{straight}\mathrm{line}. \end{gathered}$$

Required area is the shaded area OABO

$$\begin{gathered} \mathrm{Using}\mathrm{the}\mathrm{horizontal}\mathrm{strip}\mathrm{method}, \\ \mathrm{Area}\left(OABO\right)={\int }_0^3\left|x\right|dy \\ ={\int }_0^3\frac{y^2}{4}dy \\ ={\left[\frac{1}{4}\left(\frac{y^3}{3}\right)\right]}_0^3 \\ =\frac{3^3}{12} \\ =\frac{9}{4}\mathrm{sq}.\mathrm{units} \end{gathered}$$