$$\int_{ - a}^a {{{\log }_e}\left( {x + \sqrt {1 + {x^2}} } \right)dx} $$ is equal to :
A.
$$2{\log _e}a$$
B.
0
C.
$${\log _e}2 + \log \,a$$
D.
none of these
Answer :
0
Solution :
Let $$f\left( x \right) = {\log _e}\left( {x + \sqrt {1 + {x^2}} } \right).$$ Clearly, $$f\left( { - x} \right) = - f\left( { - x} \right).$$
So, $$f\left( x \right)$$ is an odd function.
$$\therefore \,I = 0$$
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-