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

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(B) We have the curve $y=4x-x^{2}$.To find the intersection points with the $x$-axis,we set $y=0$:$x(4-x)=0 \Rightarrow x=0$ or $x=4$.The required are

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$\frac{32}{3}$ sq. units

Vedclass · Answer

(B) We have the curve $y=4x-x^{2}$.To find the intersection points with the $x$-axis,we set $y=0$:$x(4-x)=0 \Rightarrow x=0$ or $x=4$.The required area is the integral of the function from $x=0$ to $x=4$:$A = \int_{0}^{4} (4x-x^{2}) dx$$A = \left[ \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{4}$$A = \left[ 2x^{2} - \frac{x^{3}}{3} \right]_{0}^{4}$$A = \left( 2(4)^{2} - \frac{(4)^{3}}{3} \right) - (0 - 0)$$A = \left( 2(16) - \frac{64}{3} \right) = 32 - \frac{64}{3}$$A = \frac{96-64}{3} = \frac{32}{3}$ sq. units.

The area of the region bounded by the curve $y=4x-x^{2}$ and the $x$-axis is

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