Interpreting average value within an integral

Question

I need help understanding what constitutes a constant in a particular integral. Consider the Mean Value Theorem for Integrals applied to a continuous function $f'(t)$ on $[a,x]$. Then $$\int_{a}^{x} f'(t) dt =f'(c)(x-a)$$ for some value $c$ in $(a,x)$. Using the Fundamental Theorem of Calculus on the left side, I rewrite the equation. $$f(x)=f(a)+f'(c)(x-a)$$ My question is what should I do with the $f'(c)$ term if I integrate again on both sides? $$\int_{a}^{x}f(t)dt=\int_{a}^{x}(f(a)+f'(c)(x-a))dt$$ Is $f'(c)$ a constant with respect to $t$? It seems to me that $t=c$ changes only when $x$ changes (assuming $a$ is fixed) but remains constant with respect to the integration variable $t$. My thinking is that when I apply the definite integral, the limits of integration are essentially fixed and $y=f'(c)$ is a horizontal line for any given limits. I'm told this is incorrect but I don't understand why.

Answer 1

Suppose that $f$ is derivable on an interval $[a,b]$. You have that

$$f(x)=f(a)+f'(c_x)(x-a)$$

for every $x\in [a,b]$. I write $c_x$ instead of $c$ because $c$ depends on $x$ (for each interval $[a,x]$ there's a different intermediate value $c_x$). Now, if you want to integrate with respect to $t$, substitute $x=t$ in the above expression:

$$f(t)=f(a)+f'(c_t)(t-a).$$

Integrating

$$\int_a^x f(t)dt=\int_a^xf(a)+f'(c_t)(t-a)dt.$$

So $c_t$ depends on the variable of integration.