(B) To find the area bounded by the curve $y=2x-x^2$ and the $x$-axis,we first find the intersection points with the $x$-axis by setting $y=0$. $2x-x^

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(B) To find the area bounded by the curve $y=2x-x^2$ and the $x$-axis,we first find the intersection points with the $x$-axis by setting $y=0$. $2x-x^2=0 \Rightarrow x(2-x)=0$,which gives $x=0$ and $x=2$. The area $A$ is given by the integral: $A = \int_0^2 (2x-x^2) dx$ $A = [x^2 - \frac{x^3}{3}]_0^2$ $A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$ $A = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3} \text{ sq units}$.

Class 12 Mathematics Application of Integration Area bounded by region of single curve

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

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A
$\frac{2}{3} \text{ sq units}$
B
$\frac{4}{3} \text{ sq units}$
C
$\frac{5}{3} \text{ sq units}$
D
$\frac{8}{3} \text{ sq units}$

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Application of Integration

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Area bounded by region of single curve

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

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