The area (in sq. units) of the region bounded | vedclass

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(A) Step $1$: Find the points where the curve $y = 2x - x^2$ intersects the $X$-axis by setting $y = 0$.$2x - x^2 = 0 \implies x(2 - x) = 0$. Thus, $x

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) Step $1$: Find the points where the curve $y = 2x - x^2$ intersects the $X$-axis by setting $y = 0$.$2x - x^2 = 0 \implies x(2 - x) = 0$. Thus, $x = 0$ and $x = 2$.Step $2$: The area $A$ is given by the integral of $y$ with respect to $x$ from $0$ to $2$.$A = \int_{0}^{2} (2x - x^2) \, dx$.Step $3$: Evaluate the integral.$A = [x^2 - \frac{x^3}{3}]_{0}^{2}$.Step $4$: Substitute the limits.$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3}) = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} 	ext{ sq. units}$.

The area (in sq. units) of the region bounded by the curve $y = 2x - x^2$ and the $X$-axis is:

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