The area (in sq. units) of the region bounded | vedclass

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(D) To find the area bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$,we first find their points of intersection by setting $x \log x = 2x - 2

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$\frac{7}{12}$

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(D) To find the area bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$,we first find their points of intersection by setting $x \log x = 2x - 2x^2$. For $x > 0$,we divide by $x$: $\log x = 2 - 2x$,which implies $\log x + 2x - 2 = 0$. By inspection,$x = 1$ is a solution since $\log(1) + 2(1) - 2 = 0 + 2 - 2 = 0$. For $x \in (0, 1]$,the curve $y = 2x - 2x^2$ lies above $y = x \log x$. The area $A$ is given by the integral $\int_{0}^{1} (2x - 2x^2 - x \log x) dx$. Evaluating the integral: $\int_{0}^{1} 2x dx = [x^2]_{0}^{1} = 1$. $\int_{0}^{1} 2x^2 dx = [\frac{2}{3}x^3]_{0}^{1} = \frac{2}{3}$. Using integration by parts for $\int x \log x dx$: $\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$. Evaluating from $0$ to $1$: $[\frac{x^2}{2} \log x - \frac{x^2}{4}]_{0}^{1} = (0 - \frac{1}{4}) - (0) = -\frac{1}{4}$. Thus,$A = 1 - \frac{2}{3} - (-\frac{1}{4}) = 1 - \frac{2}{3} + \frac{1}{4} = \frac{12 - 8 + 3}{12} = \frac{7}{12}$.

The area (in sq. units) of the region bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$ is

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