Integral of a function's derivative does not equal the original function?

I am struggling with assessing the validity of this statement.

$$\int ^{x}_{a}f'\left( t\right) dt \neq f\left( x\right) $$

I can understand that the left side yields a class of functions $F(x)$ whose derivative is $f(x)$, but doesn't that mean that the left side evaluates to $f(x) + C$ and that constant pairs with whatever constant exists in the original $f(x)$? For example, if $f(x)=2x+6$ then the antiderivative is $2x + C$ but $C$ here is just $6$, right? So doesn't the equality hold?

Solution 1:

Assuming $f$ is differentiable, then the fundamental theorem of calculus says

$$\int_3^xf'(s) \, ds=f(x)-f(3)$$

Hence, unless $f(3)=0$, the integral expression is not $f(x)$.

Solution 2:

Actually, I disagree with your statement 'the left side yields...'

You are talking about indefinite integrals, but here you have a definite integral. In particular, you have $$ \int_3^x f'(x)\,dx=g(x)-g(3), $$ where $g$ is any antiderivative of $f'(x)$. In particular, we know that all antiderivatives of $f'(x)$ are of the form $f(x)+C$ for some constant $C$, so that $$ \int_3^x f'(x)\,dx=[f(x)+C]-[f(3)+C]=f(x)-f(3). $$ So, your question boils down to this: is $f(x)=f(x)-f(3)$ true for all $x$? The answer will depend on the particular value that your function $f$ assigns to the input $3$.