Question

If $f\left( x \right)$ is a twice differentiable function on the interval$a < x < b$, and $f^{‘}(x)=c$ for all$a < x < b$, then $\int_a^b {f''\left( x \right)} dx =$
A) $b - a$
B) $c\left( {b - a} \right)$
C) 0
D) $c\left( {a - b} \right)$
E) $a - b$

Answer 1

Here we will first find the value of $f''\left( x \right)$ by differentiating $f'\left( x \right) = c$ with respect to $x$. Here, $c$ is a constant, therefore, it is a constant function. We will use the fact that the differentiation of any constant function is zero. The differentiation of any constant function is zero and the integration of zero is zero.

Complete step by step solution:
Let $I = \int_a^b {f''\left( x \right)} dx$……..$\left( 1 \right)$
We will find the value of $f''\left( x \right)$ now. It is given that here $c$ is a constant. That means, $f'\left( x \right) = c$ is a constant function.
$\Rightarrow f'\left( x \right) = c$ ……..$\left( 2 \right)$
We will differentiate the equation $\left( 1 \right)$ with respect to $x$ on both sides.
Therefore,
$\Rightarrow \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = \dfrac{{dc}}{{dx}}$ …………$\left( 3 \right)$
Here $c$ is a constant and we know the differentiation of any constant with respect to any variable is zero. Therefore, the right side of the equation will become zero. So,
$\Rightarrow f''\left( x \right) = 0$
Putting value of $f''\left( x \right)$ in equation $\left( 1 \right)$, we get
$I = \int_a^b 0 dx$
The right becomes zero as the integration of zero is zero.
$I = 0$
Thus, the value of $\int_a^b {f''\left( x \right)} dx$ is zero.

Hence, the correct option is C.

Note:
Here, we have used both differentiation and integration to find the value. Differentiation represents the rate of change of a function whereas integration represents an accumulation or sum of a function over a range. They are inverse of each other. We have not substituted the limit because the limit is substituted if there is any variable left after the integration of the function. As, the answer is 0, that means no variable is present, so limit is not substituted.