(B) According to the Fundamental Theorem of Calculus,$\int_1^2 f'(x) dx = [f(x)]_1^2 = f(2) - f(1)$.Since $f(x)$ satisfies the conditions of Rolle's t

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(B) According to the Fundamental Theorem of Calculus,$\int_1^2 f'(x) dx = [f(x)]_1^2 = f(2) - f(1)$.Since $f(x)$ satisfies the conditions of Rolle's theorem on the interval $[1, 2]$,one of the necessary conditions is that $f(1) = f(2)$.Therefore,$f(2) - f(1) = 0$.Hence,$\int_1^2 f'(x) dx = 0$.

Class 12 Mathematics Continuity and Differentiation Rolle’s theorem, Lagrange's mean value theorem

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If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$,then $\int_1^2 f'(x) dx$ is equal to

A
$3$
B
$0$
C
$1$
D
$2$

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Continuity and Differentiation

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Rolle’s theorem, Lagrange's mean value theorem

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MediumWBJEE 2025

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MediumWBJEE 2018

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If $f(x) = \cos x$ for $0 \le x \le \frac{\pi}{2}$,then the real number $c$ of the Mean Value Theorem is:

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If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$,then $\int_1^2 f'(x) dx$ is equal to

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