If (alpha, beta) is the stationary point of t | vedclass

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(C) Given the curve $y=2x-x^2$. For the stationary point,we set $\frac{dy}{dx}=0$. $\frac{dy}{dx} = 2-2x = 0 \Rightarrow x=1$. Thus,$\alpha = 1$. The

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$\frac{3-\log 4}{3 \log 2}$

Vedclass · Answer

(C) Given the curve $y=2x-x^2$. For the stationary point,we set $\frac{dy}{dx}=0$. $\frac{dy}{dx} = 2-2x = 0 \Rightarrow x=1$. Thus,$\alpha = 1$. The area is bounded by $y=2^x, y=2x-x^2, x=0$ and $x=1$. Required Area $= \int_0^1 (2^x - (2x-x^2)) dx$. $= \int_0^1 2^x dx - \int_0^1 (2x-x^2) dx$. $= \left[ \frac{2^x}{\log 2} \right]_0^1 - \left[ x^2 - \frac{x^3}{3} \right]_0^1$. $= \left( \frac{2^1}{\log 2} - \frac{2^0}{\log 2} \right) - \left( (1^2 - \frac{1^3}{3}) - (0) \right)$. $= \frac{2-1}{\log 2} - (1 - \frac{1}{3}) = \frac{1}{\log 2} - \frac{2}{3}$. $= \frac{3 - 2 \log 2}{3 \log 2} = \frac{3 - \log 4}{3 \log 2}$.

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

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