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

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(C) The region is bounded by $y = 2^x$ and $y = x + 1$ (since $x \ge 0$ in the first quadrant,$|x + 1| = x + 1$) from $x = 0$ to $x = 1$.The required

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$\frac{3}{2} - \frac{1}{\log_e 2}$

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(C) The region is bounded by $y = 2^x$ and $y = x + 1$ (since $x \ge 0$ in the first quadrant,$|x + 1| = x + 1$) from $x = 0$ to $x = 1$.The required area is given by the integral:$A = \int_{0}^{1} ((x + 1) - 2^x) dx$Evaluating the integral:$A = \left[ \frac{x^2}{2} + x - \frac{2^x}{\ln 2} \right]_{0}^{1}$Substituting the limits:$A = \left( \frac{1^2}{2} + 1 - \frac{2^1}{\ln 2} \right) - \left( \frac{0^2}{2} + 0 - \frac{2^0}{\ln 2} \right)$$A = \left( \frac{1}{2} + 1 - \frac{2}{\ln 2} \right) - \left( 0 + 0 - \frac{1}{\ln 2} \right)$$A = \frac{3}{2} - \frac{2}{\ln 2} + \frac{1}{\ln 2}$$A = \frac{3}{2} - \frac{1}{\ln 2}$

The area (in sq. units) of the region bounded by the curves $y = 2^x$ and $y = |x + 1|$ in the first quadrant is

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