Mathematics | MHT CET 2016 | Discussion Page

The area of the region bounded by the curve y=2x-x² and X- axis is

$\frac{2}{3}$ sq units

$\frac{4}{3}$ sq units

$\frac{5}{3}$ sq units

$\frac{8}{3}$ sq units

Option B

Given equation of curve is y=2x-x²

$\Rightarrow$ $x^{2}-2x=-y$

$\Rightarrow$ $x^{2}-2x+1$ = -y+1

$\Rightarrow$ $(x-1)^{2}$ =-(y-1)

This is the equation of parabola having vertex (1,1) and open downward.

The parabola intersect the X-axis , put y=0 , we get

$0=2x-x^{2}$

$\Rightarrow$ x(2-x)=0

$\Rightarrow$ x=0,2

$\therefore$ Area of bounded region between the curve and X- axis

$=\int_{0}^{2} y dx$

$=\int_{0}^{2} (2x-x^{2}) dx=\left[\frac{2x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}$

$=\left[4-\frac{8}{3}-0-0\right]=\frac{4}{3}$ sq. units

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