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

The area of the region bounded by the curves $y=\mathrm{e}^x, y=\log x$ and lines $x=1, x=2$ is
  1. $(\mathrm{e}-1)^2 \mathrm{sq}$. units
  2. $\left(\mathrm{e}^2-\mathrm{e}+1\right) \mathrm{sq}$. units
  3. $\left(\mathrm{e}^2-\mathrm{e}+1-2 \log 2\right) \mathrm{sq}$. units
  4. $\left(\mathrm{e}^2+\mathrm{e}-2 \log 2\right)$ sq. units

Solution

$\begin{aligned} & \text {Required Area } \\ & =\int_1^2\left(\mathrm{e}^x-\log x\right) \mathrm{d} x \\ & =\left[\mathrm{e}^x\right]_1^2-\int_1^2 1 \log x \mathrm{~d} x \\ & =\left(\mathrm{e}^2-\mathrm{e}\right)-\left[x \log x-\int_1^2 1 \mathrm{~d} x\right] \\ & =\left(\mathrm{e}^2-\mathrm{e}\right)-[x \log x-x]_1^2 \\ & =\left(\mathrm{e}^2-\mathrm{e}\right)-[(2 \log 2-2)-(1 \log 1-1)] \\ & =\mathrm{e}^2-\mathrm{e}-(2 \log 2-2-0+1) \\ & =\mathrm{e}^2-\mathrm{e}-(2 \log 2-1) \\ & =\left(\mathrm{e}^2-\mathrm{e}+1-2 \log 2\right) \text { sq. units }\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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