Why do we not include $c$ in the computation of the definite integral? Why is it when evaluating the definite integral we commonly opt to omit the constant $c$
$$\int_1^2x^2 \, dx= \left.\frac{x^3}{3} \right|_1^2 =\frac{2^3}{3}-\frac{1^3}{3}=\frac{7}{3}$$
But when evaluating the indefinite integral we include it?
$$\int x^2dx = \frac{x^3}{3} + c$$
Would it not be more accurate to instead do the following?
$$\int_1^2 x^2\,dx=\left.\frac{x^3}{3}+c\right|_1^2 = \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \left(\frac{7}{3}+c\right)$$
Is there a historical reason to all of this?
 A: Besides the algebraic fact that in your computation it should be $c-c=0$, you are missing the point of what an integral means: the symbol 
$$
\int_1^2x^2\,dx
$$
denotes a number (7/3, in this case), so there is no place for a constant. 
A: Because it is a definite integral. The indefinite integral is simply an antiderivative:
$$ \int f'(x)\,dx = f(x) + C
$$
For any $C$, $f(x) + C$ is an antiderivative of $f'(x)$.
The definite integral associates a number with an interval and a function. That is,
$$ I = \int_a^b f'(x)\,dx
$$
These are two different things, so there is no reason to include $C$ in a definite integral.
The above two concepts are connected by the fundamental theorem of calculus, which states that
$$ I = \int_a^b f'(x)\,dx = \phi(b) - \phi(a)
$$
where $\phi(x)$ is any antiderivative of $f'(x)$. Obviously $\phi(x) = f(x) + C$ would work, and the $C$s cancel each other out, as desired.
A: It might be useful to consider graphically what the constant of integration means. When you take the derivative of a constant, you know it vanishes. 
Start with a function, call it $f(x)$. Call its antiderivative, $F(x) = ∫ f(x) dx$. 
The graph of $F(x)$ could be moved up and down along the y-axis, any position would work. The slope of $F(x)$, its derivative $F'(x)$, is always the same no matter how high you slide it up the y-axis, or how low it is. 
The definite integral is evaluating the relative difference between two points on the curve $F(x)$, $F(a)-F(b)$. No matter how high or low $F(x)$ is, the relative difference between $F(a)$ and $F(b)$ remains the same. It does not matter what the constant of integration is, just the relative difference of the points on the antiderivative. 
The curve could be shifted up or down arbitrarily (that is what the constant of integration represents), but the "canceling" out of the constant is graphically apparent in that we are finding a relative difference, not an absolute one. 
I hope that helped a bit. By the way, if you want to actually try out an example, my favorite is $∫ x^2 dx = \frac{1}{3} x^3 +C$
Notice all of the things above by trying to shift the graph up or down, and see how the constant of integration doesn't matter for finding relative differences of function values on the curve, nor does it matter for differentiation. 
A: $$
\text{Becuase } \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \frac{7}{3} \ne \left(\frac{7}{3}+c\right).
$$
A: One simplistic way of looking at it is to say that the constant cancels out when you evaluate the definite integral. You can see that from your own question, if you do the algebra properly.
But there's a deeper answer: the indefinite integral needs the constant there because it's not a single function, but rather an entire family of functions.
It's possible to evaluate a single indefinite integral in two different ways to get quite different answers if you neglect the constant.
As an example, consider $I = \int\frac{\sin 2x}{1 + \cos 2x} dx$ I'm going to ignore the constant of integration for instructive purposes.
One "obvious" way to do it is to make the substitution $u = 1 + \cos 2x$, giving the answer $I = -\frac 12\ln(1+\cos 2x)$. You can use the double angle formula to simplify that to $I =-\ln(\cos x) -\frac 12\ln 2$.
Another only slightly less obvious way is to immediately apply the double angle formulae to the integrand (both numerator and denominator). This will make the integral $I = \int \frac{2\sin x \cos x}{2\cos^2 x}dx = \int \tan x dx = -\ln(\cos x)$.
You should be able to see immediately that the answers are different by that $-\frac 12 \ln 2$ term. The indefinite integral gives you an answer that is actually a infinite set of functions that differ from one another by a constant (or, in more advanced multivariable integrals, by a function independent of the variable of integration). To be mathematically correct, you need that constant at the end (or an arbitrary function in the case of the multivariable form, but perhaps you shouldn't worry about that now).
The definite integral is different in the sense that it evaluates either to a number (in the single variable case) or a function independent of the variable of integration (in the multivariable case). It is used for more "practical" or "physical" applications such as finding the area under the curve or the distance a particle with continuously varying speed travels over a given time interval. Clearly, there's just one correct answer for such problems, and hence the definite integral does not require an arbitrary constant.
