First slide

Area of bounded Regions

Question

$Let f : R^{+} \to R be a differentiable function satisfying$ $f (x) = e + (1 - x) \ln (\frac{x}{e}) + \int_{1}^{x} f (t) d t, for all x ε R^{+} . If the area enclosed by the curve$ $g (x) = x (f (x) - e^{x})$ $l y i n g i n t h e f o u r t h q u a d r a n t i s A, t h e n A =$

Difficult

A

$\frac{1}{2}$
B

$\frac{1}{3}$
C

$\frac{1}{4}$
D

$\frac{2}{3}$

Solution

$f^{'} (x) = 0 + (1 - x) (\frac{1}{x}) + \ln (\frac{x}{e}) (- 1) + f (x) (b y n e w t o n' s L e i b n i t z r u l e)$
$\Rightarrow f^{'} (x) = \frac{1}{x} - 1 + (\ln x - 1) (- 1) + f (x)$
$\Rightarrow \frac{d y}{d x} - y = \frac{1}{x} - \ln x$
$\Rightarrow y e^{- x} = \int (\frac{1}{x} - \ln x) e^{- x} d x (s o l v i n g o f l i n e a r d i f f e r e n t i a l e q u a t i o n)$
$\Rightarrow y e^{- x} = e^{- x} (\log x) + c$
$\sin c e f (1) = e \Rightarrow e e^{- 1} = 0 + c \Rightarrow c = 1$
$y = \log x + e^{x}$
$f (x) = e^{x} + \ln x \Rightarrow g (x) = x (f (x) - e^{x}) = x \ln x$
$A r e a = A = \int_{0}^{1} x \ln x d x = \frac{1}{4}$

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