If $$f\left( x \right)$$ satisfies the conditions of Rolle’s theorem in [1, 2] then $$\int_1^2 {f'\left( x \right)} dx$$ is equal to :
A.
1
B.
3
C.
0
D.
none of these
Answer :
0
Solution :
As $$f\left( x \right)$$ satisfies the conditions of Rolle’s theorem in [1, 2], $$f\left( x \right)$$ is continuous in the interval and $$f\left( 1 \right) = f\left( 2 \right).$$
$$\therefore \int_1^2 {f'\left( x \right)} dx = \left[ {f\left( x \right)} \right]_1^2 = f\left( 2 \right) - f\left( 1 \right) = 0$$
Releted MCQ Question on Calculus >> Definite Integration
Releted Question 1
The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-