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? 
 A: 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 $f(x)$.
A: 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 $3$.
A: In addition to the excelent answers already given, there are a few subleties one should explicitly point out.
There are two main concepts for integration, the first being indefinite integration, that is finding the antiderrivative of a function, and the second is definite integration, finding the measure of the (signed) area enclosed by the graph of a function and the x-axis. There are various ways to codify these concepts in a rigourous mathematical language.
For exaple, if we know that $f$ is a real function over $[a, b]$, $a,b \in \mathbb{R}$, we use Riemann's definition of
$$\int_{a}^b f(x) \, \operatorname{d}\!x \ ,$$
which can be found in any elementary calculus textbook. Provited that $f$ meets some specific conditions, we say that $f$ is Riemann-integrable over $[a,b]$ and we assign the above symbolic expression a unique real value. (There is also a theorem according to which if $f$ is Riemann-integrale over $[a,b]$, it's also Riemann-integrable over any closed subinterval of $[a,b]$).
But what about indefinite integration? Remember that, given a real function $h$ over an interval $I$, function $H:I \to \mathbb{R}$ is called an antiderivative of $h$ just in case $H'=h$. You shall note that whenever $H$ is an antiderivative of $h$, $H+c$ is likewise an antiderivative of $h$, and you can prove that any antiderivative of $h$ is of the form $H+c$ for some constant c.
Now, get ready for the hard truth...
While acknowledging that there is an infinite number of antiderivatives for $h$ (all functions $H+c$), mathematicians go mad, break their own rules, and proudly state:
$$\int h(x) \, \operatorname{d}\!x = H(x),$$
without any clue of shame! Sometimes, we even go as far as claiming that ‘the indefinite integral of $h(x)$ is H(x) for all $x \in I$’! There is not really such a thing as the indefinite integral of a function, for there is not such a thing as the antiderivative of a function; there are uncountably many of them. This is a clear abuse of notation/terminology, but due to historical reasons, it is accepted as the standard to this day.
To illustrate the above mentioned facts, some books prefer to write
$$\int h(x) \, \operatorname{d}\!x = H(x)+c,$$
and say that $c$ is an arbitrary constant, named constant of integration. This is a good way to remind us that if we have an integral equation like this:
$$\int h(x) \, \operatorname{d}\!x + x^2 = 6\ln x + 42,$$
we get the equivalent statement
$$\exists c \in \mathbb{R} \quad H(x) + c + x^2 = 6\ln x +42.$$
$$\star \ \star \ \star$$
Now you may understand what exactly that peculiar constant is, and why it's not supposed to much a specific function, as you say in your question. There are not constants ‘existing’ in integrable functions!
I'll finish this long (too long?) post by noticing that the integral
$$\int_{a}^x f'(t) \, \operatorname{d}\!t$$
in question is not just a definite integral (definite integrals equal a real number) but a function
$$F(x) = \int_{a}^x f'(t) \, \operatorname{d}\!t, \ x \in D,$$
where $f'$ should be Riemann-integrable over $D$ (recall also that $a \in D$, and $D$ should be a closed and bounted interval). After all this, you can use the Fundamental Theorem of Calculus and find that
$$F(x) = f(x) - f(a), \ x \in D,$$
as already mentioned by others.
I know I wrote a lot, but I think this helps the OP figure out some usual misconceptions that underlie her/his question. Any comments/corrections welcome!
