The area (in sq. units) enclosed by the curve | vedclass

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(A) To find the area enclosed by the curves $y=2x-x^2$ and $y=x^2-2x-6$, we first find their points of intersection by setting the equations equal to

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$\frac{64}{3}$

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(A) To find the area enclosed by the curves $y=2x-x^2$ and $y=x^2-2x-6$, we first find their points of intersection by setting the equations equal to each other:$2x-x^2 = x^2-2x-6$$2x^2-4x-6 = 0$$x^2-2x-3 = 0$$(x-3)(x+1) = 0$Thus, the points of intersection are at $x=-1$ and $x=3$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x=-1$ to $x=3$:$A = \int_{-1}^{3} [(2x-x^2) - (x^2-2x-6)] dx$$A = \int_{-1}^{3} (-2x^2+4x+6) dx$$A = [- \frac{2}{3}x^3 + 2x^2 + 6x]_{-1}^{3}$Evaluating at the limits:$A = [(- \frac{2}{3}(27) + 2(9) + 6(3)) - (- \frac{2}{3}(-1) + 2(1) + 6(-1))]$$A = [(-18 + 18 + 18) - (\frac{2}{3} + 2 - 6)]$$A = 18 - (\frac{2}{3} - 4) = 18 - (\frac{2-12}{3}) = 18 - (-\frac{10}{3}) = 18 + \frac{10}{3} = \frac{54+10}{3} = \frac{64}{3}$ sq. units.

The area (in sq. units) enclosed by the curves $y=2x-x^2$ and $y=x^2-2x-6$ is

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