First slide

Area of bounded Regions

Question

$Let the functions f : R \to R and g : R \to R be defined by$ $f (x) = e^{x - 1} - e^{- | x - 1 |} and g (x) = \frac{1}{2} (e^{x - 1} + e^{1 - x})$
$Then the area of the region in the first quadrant bounded by the curves y = f (x), y = g (x) and x = 0 is$

Difficult

A

$(2 - \sqrt{3}) + \frac{1}{2} (e - e^{- 1})$
B

$(2 + \sqrt{3}) + \frac{1}{2} (e - e^{- 1})$
C

$(2 - \sqrt{3}) + \frac{1}{2} (e + e^{- 1})$
D

$(2 + \sqrt{3}) + \frac{1}{2} (e + e^{- 1})$

Solution

here,
$\begin{aligned} f (x) = \{\begin{array}{cc} 0 & x \leq 1 \\ e^{x - 1} - e^{1 - x} & x \geq 1 \end{array} \\ & g (x) = \frac{1}{2} (e^{x - 1} + e^{1 - x}) \end{aligned}$
$\begin{array}{l} solve f (x) & g (x) t h e r e f o r e e^{x - 1} - e^{1 - x} = \frac{1}{2} (e^{x - 1} + e^{1 - x}) \Rightarrow e^{2 x - 2} = 3 \Rightarrow x = 1 + l n \sqrt{3} \\ So bounded area = \int_{0}^{1} \frac{1}{2} (e^{x - 1} + e^{1 - x}) dx + \int_{1}^{1 + \ln \sqrt{3}} [\frac{1}{2} (e^{x - 1} + e^{1 - x}) - (e^{x - 1} + e^{1 - x})] dx \\ = \frac{1}{2} {[e^{x - 1} - e^{1 - x}]}_{0}^{1} + {[- \frac{1}{2} e^{x - 1} - \frac{3}{2} e^{1 - x}]}_{1}^{1 + \ln \sqrt{3}} \\ = \frac{1}{2} [e - \frac{1}{e}] + [(- \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}) + 2] = 2 - \sqrt{3} + \frac{1}{2} (e - \frac{1}{e}) \end{array}$

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