The area bounded by the curve y = e^x and the | vedclass

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(C) The area is bounded by $y = e^x$ and $y = |x - 1|$ between $x = 0$ and $x = 2$.Since $|x - 1| = 1 - x$ for $x \in [0, 1]$ and $|x - 1| = x - 1$ fo

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$e^2 - 2$

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(C) The area is bounded by $y = e^x$ and $y = |x - 1|$ between $x = 0$ and $x = 2$.Since $|x - 1| = 1 - x$ for $x \in [0, 1]$ and $|x - 1| = x - 1$ for $x \in [1, 2]$,the area is given by:Area $= \int_{0}^{1} (e^x - (1 - x)) dx + \int_{1}^{2} (e^x - (x - 1)) dx$$= \int_{0}^{1} (e^x - 1 + x) dx + \int_{1}^{2} (e^x - x + 1) dx$$= [e^x - x + \frac{x^2}{2}]_{0}^{1} + [e^x - \frac{x^2}{2} + x]_{1}^{2}$$= (e^1 - 1 + \frac{1}{2}) - (e^0 - 0 + 0) + (e^2 - \frac{4}{2} + 2) - (e^1 - \frac{1}{2} + 1)$$= (e - \frac{1}{2}) - 1 + (e^2 - 2 + 2) - (e + \frac{1}{2})$$= e - \frac{1}{2} - 1 + e^2 - e - \frac{1}{2}$$= e^2 - 2$

The area bounded by the curve $y = e^x$ and the lines $y = |x - 1|, x = 0, x = 2$ is given by

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