SATHEE: Area Question 30

Area Question 30

Question 30

Area of the region bounded by the curve $y = e^{x}$ and lines $x = 0$ and $y = e$ is

(2009) (a) $e - 1$ (b) $\int_{1}^{e} \ln (e + 1 - y) d y$ (c) $e - \int_{0}^{1} e^{x} d x$ (d) $\int_{1}^{e} \ln y d y$

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Solution:

Shaded area $= e - \int_{0}^{1} e^{x} d x = 1$

Also, $\int_{1}^{e} \ln (e + 1 - y) d y$ [put $e + 1 - y = t \Rightarrow - d y = d t$ ]

$= \int_{e}^{1} \ln t (- d t) = \int_{1}^{e} \ln t d t = \int_{1}^{e} \ln y d y = 1$

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