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The area of the region bounded by $y=2x-x^2$ and x-axis is : |
$\frac{2}{3}$ sq.units $\frac{4}{3}$ sq.units $\frac{4}{5}$ sq.units $\frac{5}{3}$ sq.units |
$\frac{4}{3}$ sq.units |
The correct answer is Option (2) → $\frac{4}{3}$ sq.units $y=2x-x^2-1+1$ $(y-1)=-(x-1)^2$
curve intersects x axis at y = 0 $⇒2x=x^2$ so $x=0,2$ Required area $=\int\limits_0^22x-x^2dx$ $=\left[x^2-\frac{x^3}{3}\right]_0^2$ $=\frac{4}{3}$ sq.units |
