Let $$f\left( x \right) = \frac{{{e^x} + 1}}{{{e^x} - 1}}$$ and $$\int_0^1 {\frac{{{e^x} + 1}}{{{e^x} - 1}}.x\,dx} = \lambda .$$ Then $$\int_{ - 1}^1 {tf\left( t \right)dt} $$ is equal to :
A.
0
B.
$$2\lambda $$
C.
$$\lambda $$
D.
none of these
Answer :
$$2\lambda $$
Solution :
$$\eqalign{
& \phi \left( t \right) = tf\left( t \right) = t.\frac{{{e^t} + 1}}{{{e^t} - 1}} \cr
& {\text{So, }}\phi \left( { - t} \right) = - t.\frac{{{e^{ - t}} + 1}}{{{e^{ - t}} - 1}} = t.\frac{{{e^t} + 1}}{{{e^t} - 1}} \cr
& \therefore \phi \left( t \right) = \phi \left( { - t} \right){\text{.}}\,{\text{Hence, }}\phi \left( t \right){\text{ is an even function}} \cr
& \therefore \int_{ - 1}^1 {\phi \left( t \right)dt} = 2\int_0^1 {\phi \left( t \right)dt} = 2\lambda \cr} $$
Releted MCQ Question on Calculus >> Application of Integration
Releted Question 1
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-