(C) The required area is given by the integral of the upper curve minus the lower curve between the limits $x=1$ and $x=2$. $\text{Area} = \int_1^2 (e

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$(e^2-e+1-2 \log 2) \text{ sq. units}$

(C) The required area is given by the integral of the upper curve minus the lower curve between the limits $x=1$ and $x=2$. $\text{Area} = \int_1^2 (e^x - \log x) dx$ $= \int_1^2 e^x dx - \int_1^2 \log x dx$ $= [e^x]_1^2 - [x \log x - x]_1^2$ $= (e^2 - e^1) - [(2 \log 2 - 2) - (1 \log 1 - 1)]$ Since $\log 1 = 0$,we have: $= e^2 - e - [2 \log 2 - 2 - 0 + 1]$ $= e^2 - e - [2 \log 2 - 1]$ $= e^2 - e + 1 - 2 \log 2 \text{ sq. units}$

Class 12 Mathematics Application of Integration Area bounded by region of multi curve

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The area of the region bounded by the curves $y=e^x, y=\log x$ and lines $x=1, x=2$ is

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A
$(e-1)^2 \text{ sq. units}$
B
$(e^2-e+1) \text{ sq. units}$
C
$(e^2-e+1-2 \log 2) \text{ sq. units}$
D
$(e^2+e-2 \log 2) \text{ sq. units}$

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The area of the region bounded by the curves $y=e^x, y=\log x$ and lines $x=1, x=2$ is

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