y}\right\}$$

The graph of $$f$$ and $$g$$ can be plotted in the first quadrant as given below.

BITSAT 2023 Mathematics - Area Under The Curves Question 4 English Explanation

The point of intersection, $$A$$ is given by $$f(x)=g(x)$$

$$\begin{aligned} & \Rightarrow \quad e^{x-1}-e^{1-x}=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right) \\ & \Rightarrow \quad e^{x-1}=3 e^{1-x} \\ & \Rightarrow \quad\left(e^{x-1}\right)^2=3 \Rightarrow e^{2 x-2}=3 \\ & \Rightarrow \quad 2 x-2=\ln 3 \\ & \Rightarrow \quad x=1+\frac{1}{2} \ln 3 \\ & \therefore \text { Required area }=\int_\limits0^1\left(\frac{1}{2} e^{x-1}+e^{1-x}\right) d x +\int_\limits1^{1+\frac{1}{2} \ln 3}\left[\left(\frac{1}{2} e^{x-1}+e^{1-x}\right)-\left(e^{x-1}-e^{1-x}\right)\right] d x \end{aligned}$$

$$\begin{aligned} & =\frac{1}{2}\left[e^{x-1}-e^{1-x}\right]_0^1-\left[\frac{1}{2} e^{x-1}+\frac{3}{2} e^{1-x}\right]_1^{1+\frac{1}{2} \ln 3} \\ & =\frac{1}{2}\left(0-\frac{1}{e}+e\right)-\frac{1}{2}\left[e^{\ln \sqrt{3}}+3 e^{-\ln \sqrt{3}}-4\right] \\ & =\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(\sqrt{3}+\frac{3}{\sqrt{3}}-4\right) \\ & =-\frac{1}{2}(2 \sqrt{3}-4)+\frac{1}{2}\left(e-\frac{1}{e}\right) \\ & =(2-\sqrt{3})+\frac{1}{2}\left(e-\frac{1}{e}\right) \end{aligned}$$

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