Question

The area bounded by the curve y = 4x − x² and the x-axis is
(a) $\frac{30}{7}$sq. units

(b) $\frac{31}{7}$sq. units

(c) $\frac{32}{3}$sq. units

(d) $\frac{34}{3}$sq. units

Answer 1

$\left(\mathrm{c}\right)\frac{32}{3}\mathrm{sq}.\mathrm{units}$


Point of intersection of parabola y = 4x − x² with x-axis is given by
$$\begin{gathered} y=4x-x^2\mathrm{and}y=0\left(\mathrm{Equation}\mathrm{of}\mathrm{x}\mathrm{axis}\right) \\ \Rightarrow 4x-x^2=0 \\ \Rightarrow x=0\mathrm{or}x=4 \\ \Rightarrow y=0,y=0 \\ \mathrm{Thus}O\left(0,0\right)\mathrm{and}B\left(4,0\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{parabola}\mathrm{and}x-\mathrm{axis}. \\ \mathrm{Required}\mathrm{shaded}\mathrm{area}={\int }_0^4\left(4x-x^2\right)dx \\ ={\left[2x^2-\frac{x^3}{3}\right]}_0^4 \\ =2\times 16-\frac{64}{3}-0 \\ =\frac{96-64}{3} \\ =\frac{32}{3}\mathrm{square}\mathrm{units} \end{gathered}$$