If $(\alpha, \beta)$ is the stationary point of the curve $y=2 x-x^2$, then the area bounded by the curves $y=2 x, y=2 x-x^2, x=0$ and $x=\alpha$ is

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Aicharya · Accepted Answer

$y=2 x-x^2$ For stationary point $\frac{d y}{d x}=0$ $\Rightarrow 2-2 x=0 \Rightarrow x=1=\alpha$
Given curves are $y=2^{\mathrm{x}}, y=2 x-x^2, x=0, x=1$

$\begin{aligned} & \text { Required area }=\int_0^1 2^x d x-\int_0^1\left(2 x-x^2\right) d x \\ & =\left[\frac{2^x}{\log 2}\right]_0^1-\left[x^2-\frac{x^3}{3}\right]_0^1=\frac{1}{\log 2}-\frac{2}{3}=\frac{3-\log 4}{3 \log 2}\end{aligned}$

If $(\alpha, \beta)$ is the stationary point of the curve $y=2 x-x^2$, then the area bounded by the curves…

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