$$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$$
2
AIEEE 2003
MCQ (Single Correct Answer)
Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is
A
$$e + {{{e^2}} \over 2} + {5 \over 2}$$
B
$$e - {{{e^2}} \over 2} - {5 \over 2}$$
C
$$e + {{{e^2}} \over 2} - {3 \over 2}$$
D
$$e - {{{e^2}} \over 2} - {3 \over 2}$$
Explanation
Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$
Integrating log
$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
The area bounded by the curves $$y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right|$$ and $$y = \left| {\ln \left| x \right|} \right|$$ is
A
$$4$$sq. units
B
$$6$$sq. units
C
$$10$$sq. units
D
none of these
Explanation
First we draw each curve as separate graph
NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
Clearly the bounded area is as shown in the following figure.