On the exactness of the calculus formulas Are calculus formulas, differentiation and  integration, exact formulas or are some approximations involved? 
That is, is the value of a definite integral of a function the exact value of the (signed) area between the graph and the axis or is it only any approximation for it, and likewise for the value of the derivative and the slope of the tangent at this point.
The question arises as when introducing these notions approximations are often  central, say, via Riemann sums. It is now not clear if the definite integral is another (sometimes more viable) way to get an approximation or if it is exact and the same for derivative and slope of the tangent. 
 A: Given $a<b$ and a continuous function $f:\>[a,b]\to{\mathbb R}$ the integral
$$\int_a^b f(x)\>dx$$
is a clear cut real number, i.e., a certain element of the set ${\mathbb R}$. This number can be familiar to you, like ${7\over 13}$, $\sqrt{5}$, or $\pi$. But maybe it has never before "occurred" in mathematics.
It is the definition of this number as a limit of Riemann sums that uses approximations; but once this definition is adopted there is no question of some fishy "approximation" involved anymore.
There remains, however, the following problem: Your function $f$ can be a simple analytic expression, like $f(x):=e^{-x^2/2}$, and your scientific pocket calculator does not have the primitive of this $f$ in store, as it is not "elementary". As a consequence your calculator can only output a numerical approximation to the integral
$$\int_0^1 e^{-x^2/2}\>dx\ ,\tag{1}$$
and cannot give a value in terms of "standard" functions and constants, like $\exp$, $\cos$, $\pi$, $\sqrt{2}$, etc. Nevertheless the expression $(1)$ defines a certain real number $\xi$ exactly, i.e., to "infinitely many" decimal places.
A: 
The short answer. Yes, all of these formulas are exact in a sence that they always offer a way to get as close to the desired result as you want.
This is quite different from the approximations in general, since they are often limited in their accuracy (for example, only give upper and lower bounds for a number you want to get).
The calculus formulas can also be said to be exact by definition, meaning the formula can be used to define the object it applies to.

As an example of this difference I can give two ways of calculating the cosine:
$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$
$$\cos x \approx 2 \left(2 \left(2 \left(2 \left(2 \left(1-\frac{x^2}{2^{11}} \right)^2-1 \right)^2-1 \right)^2-1 \right)^2-1 \right)^2-1$$
The first one is exact, meaning that if we continue adding new terms to the sequence forever, we will get infinitely close to the value of $\cos x$.
Moreover, this is a definition of $\cos x$, meaning this formula is exact by definition.
The second one is only an approximation, created by repeatedly using the double angle formula and replasing the $\cos (x/2^n)$ with the first two terms of the above series. If we now try to continue doing this forever, we will eventually get $x^2/2^{\infty}=0$ inside the bracket and our approximation will fail, giving us exactly $1$ for any value of $x$.
There are other examples of course.

I think your question is really about limits. Refresh your knowledge about sequences and limits and you might find an answer to your question yourself.
Additional questions that are connected to this one:

*

*What is infinity?


*What is an irrational number?


*What is a curve?

The longer answer.
The notion of сommensurability of line segments is important. If the ratio of lengths of two lines segments is rational, then we can use one of them to measure the other. However, this is not always the case as this diagonal paradox illustrates:

So we are introduced to irrational numbers. Are they exact? Well, any whole number is exact by definition, and so are rational numbers, if we define them as a pair of whole numbers.
An irrational number can't be represented by $\frac{m}{n}$ with $m, n$ being whole numbers, but it can always be represented as $\lim_{k \to \infty} \frac{s_k}{S_k}$ with $s_k$ and $S_k$ being two sequences of whole numbers going to $\infty$ at slightly different speeds.
So basically, an irrational number describes a class of sequences, all going to the same limit.
For example:
$$\int_{0}^1 \frac{\ln x}{x-1} dx=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$
Here two sequences belong to a certain class of sequences converging to the same limit $\frac{\pi^2}{6}$. Nothing more, nothing less.

Now about tangents and areas. How do you describe a smooth curve? As a limit of a sequence of connected line segments of progressively smaller length.
So what is a tangent then? It's a straight line going through one of the line segments. If the number of segments increases to infinity, while their length becomes zero, our line has a tangent at any point (or so we like to imagine).
See the picture below for a circle.

Once we learned to define the length of any straight line segment in terms of some given line segment, we can move on to define the length of a curve and the area under it as limits.
A: Let $f(x)= x^3$, so that $f'(x)=3x^2$, and $f(2)=8$ and $f'(2)=12$.
The question would be about the number $12$: Is it exact?  When $x=2$ and $f(x)=8$, then is $f(x)$ changing exactly $12$ times as fast as $x$ is changing?
The answer is "yes".  Consider $12\pm0.000000001$  First, every number that is not between those two can be ruled out as the value of $f'(2)$ by considering values of $x$ lying close enough to $2$ (it wouldn't be hard to figure out how close is close enough but at this moment I haven't done that. That might make $12$ appear to be an approximation: we only know we're looking for a number between $12\pm0.000000001$.  But then consider $12\pm0.000000000000000000001$.  We can rule out every number not between those by focusing on an even smaller interval about $x=2$.  And so on.  It can be shown that no matter how small $\varepsilon$ is, we can rule out all numbers not between $12\pm\varepsilon$ by considering a sufficiently small interval about $x=2$.  How small is small enough depends on how small $\varepsilon$ is.  Thus no number other than $12$, no matter how close to $12$, fails to get ruled out.
Here's one way to look at how to rule out numbers other than $12$.  The point $(2,f(2)) = (2,8)$ is on the graph of the function.  Thus every line passing through that point is of the form $y-8 = m(x-12)$.  If $m$ is not exactly $12$, we want to show that it is either too big or too small to be the slope at that point.  Quite simply, if $m>12$, then there is some interval about $x=2$ within which the line $y-8=m(x-2)$ lies above the curve $y=x^3$ when $x>2$ and below the curve when $x<2$.  That at least means that $12$ is not too big to be the slope at that point.  But if $m<12$,  then there is some interval about $x=2$ within which the line $y-8=m(x-2)$ lies below the curve when $x>2$ and above the curve when $x<2$. That means $12$ is not too small to be the slope.  And one can show that ever number $>12$ is in that sense too big and every number $<12$ is in that sense too small.  Only $12$ remains as the only one not ruled out as too big or too small.
Similar things apply to integrals.
