The area bounded by the curve y = 4x - x^2 an | vedclass

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(C) Given the curve $y = 4x - x^2$ and the $x$-axis $(y = 0)$.To find the points of intersection with the $x$-axis,set $y = 0$:$4x - x^2 = 0 \implies

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq. unit}$

Vedclass · Answer

(C) Given the curve $y = 4x - x^2$ and the $x$-axis $(y = 0)$.To find the points of intersection with the $x$-axis,set $y = 0$:$4x - x^2 = 0 \implies x(4 - x) = 0 \implies x = 0, 4$.The required area is given by the definite integral:$	ext{Area} = \int_{0}^{4} (4x - x^2) dx$$= \left[ \frac{4x^2}{2} - \frac{x^3}{3} ight]_{0}^{4}$$= \left[ 2x^2 - \frac{x^3}{3} ight]_{0}^{4}$$= \left( 2(4)^2 - \frac{(4)^3}{3} ight) - (0 - 0)$$= 32 - \frac{64}{3}$$= \frac{96 - 64}{3} = \frac{32}{3} 	ext{ sq. unit}$.

The area bounded by the curve $y = 4x - x^2$ and the $x$-axis is:

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