The limit of the area under the curve y = e^{ | vedclass

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(D) The area $A$ under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ is given by the definite integral:$A = \int_{0}^{h} e^{-x} dx$To find the limit

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$1$

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(D) The area $A$ under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ is given by the definite integral:$A = \int_{0}^{h} e^{-x} dx$To find the limit as $h \rightarrow \infty$,we evaluate:$\lim_{h \rightarrow \infty} A = \lim_{h \rightarrow \infty} \int_{0}^{h} e^{-x} dx$Integrating $e^{-x}$ gives $-e^{-x}$:$= \lim_{h \rightarrow \infty} [-e^{-x}]_{0}^{h}$Applying the limits of integration:$= \lim_{h \rightarrow \infty} (-e^{-h} - (-e^{0}))$$= \lim_{h \rightarrow \infty} (-e^{-h} + 1)$Since $\lim_{h \rightarrow \infty} e^{-h} = 0$,we have:$= -0 + 1 = 1$

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

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