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

Question

KollegeApply · Accepted Answer

$\frac{9}{4}$ 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 y^2 = 4x \Rightarrow 3^2 = 4x \Rightarrow x = $\frac{9}{4}$ \text{ Thus A }\left( $\frac{9}{4}$ , 3 \right)\text{ is the point of intersection of the parabola and straight line }. Required area is the shaded area OABO Using the horizontal strip method , \text{ Area }\left( OABO \right) = \int_0^3 \left| x \right| dy = \int_0^3 y^$\frac{2}{4}$ dy = \left[ $\frac{1}{4}$\left( y^$\frac{3}{3}$ \right) \right]_0^3 = 3^$\frac{3}{12}$ = $\frac{9}{4}$\text{ sq . units } [Solution diagram: 44148_solution_1.png]

AChoose Your Answer