The area of the region bounded by the lines x | vedclass

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(B) Given the curves:$x(y-e^x)=\sin x \implies y = \frac{\sin x}{x} + e^x$$2xy = 2\sin x + x^3 \implies y = \frac{\sin x}{x} + \frac{x^2}{2}$For $x \i

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$e^2-e-\frac{7}{6}$

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(B) Given the curves:$x(y-e^x)=\sin x \implies y = \frac{\sin x}{x} + e^x$$2xy = 2\sin x + x^3 \implies y = \frac{\sin x}{x} + \frac{x^2}{2}$For $x \in [1, 2]$,$e^x > \frac{x^2}{2}$,so the area $A$ is given by:$A = \int_{1}^{2} \left( \left( \frac{\sin x}{x} + e^x \right) - \left( \frac{\sin x}{x} + \frac{x^2}{2} \right) \right) dx$$A = \int_{1}^{2} \left( e^x - \frac{x^2}{2} \right) dx$$A = \left[ e^x - \frac{x^3}{6} \right]_{1}^{2}$$A = \left( e^2 - \frac{2^3}{6} \right) - \left( e^1 - \frac{1^3}{6} \right)$$A = e^2 - \frac{8}{6} - e + \frac{1}{6}$$A = e^2 - e - \frac{7}{6}$

The area of the region bounded by the lines $x=1, x=2$,and the curves $x(y-e^x)=\sin x$ and $2xy=2\sin x+x^3$ is

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