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

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

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$\frac{4}{3}$

Vedclass · Answer

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

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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