The area of the region bounded by the curve y | vedclass

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(A) Given curves are $y=2x-x^2$ and $y=x$.To find the points of intersection,set $2x-x^2 = x$.$x-x^2 = 0 \implies x(1-x) = 0$.Thus,the points of inter

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$\frac{1}{6}$

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(A) Given curves are $y=2x-x^2$ and $y=x$.To find the points of intersection,set $2x-x^2 = x$.$x-x^2 = 0 \implies x(1-x) = 0$.Thus,the points of intersection are $x=0$ and $x=1$.The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.$	ext{Area} = \int_0^1 (y_{	ext{upper}} - y_{	ext{lower}}) dx = \int_0^1 ((2x-x^2) - x) dx$.$	ext{Area} = \int_0^1 (x-x^2) dx$.Evaluating the integral: $\left[ \frac{x^2}{2} - \frac{x^3}{3} ight]_0^1$.$= (\frac{1}{2} - \frac{1}{3}) - (0 - 0) = \frac{3-2}{6} = \frac{1}{6}$ square units.

The area of the region bounded by the curve $y=2x-x^2$ and the line $y=x$ is . . . . . . square units.

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