How can a function have an antiderivative that can't be written? If a function's integral can't be written, then how can we find exact values for it over areas? Can we only ever estimate it? Why can't we make new functions to define these strange unwritable anti-derivatives?
 A: Just because we can't write down a closed form for an antiderivative (that we know exists), that does not mean we are forced to only estimate its definite integral.
For example, there does not exist a closed form for the antiderivative $\int e^{-x^2}\,dx$, but that does not mean we are limited to merely estimating $\int_0^1 e^{-x^2}\,dx$. In fact, this is a main thrust of the theory of Taylor series!
Using $\displaystyle e^x=\sum_{n=0}^\infty {x^n\over n!}$, we see
$$
e^{-x^2}=\sum_{n=0}^\infty {(-x^2)^n\over n!}=\sum_{n=0}^\infty {(-1)^n x^{2n}\over n!}=1-x^2+{x^4\over 2!}-{x^6\over 3!}+\cdots
$$
Then, by term-by-term integration of Taylor series,
\begin{align*}
\int_0^1 e^{-x^2}\,dx&=\int_0^1 \left(1-x^2+{x^4\over 2!}-{x^6\over 3!}+\cdots\right)\,dx\\
&=\left[x-{x^3\over 3}+{x^5\over 5\cdot2!}-{x^7\over 7\cdot3!}+\cdots\right]_0^1\\
&=1-{1\over 3}+{1\over 5\cdot2!}-{1\over 7\cdot3!}+\cdots
\end{align*}
Thus, we have found the exact value---not an estimate---for the desired area under the curve even though no closed form antiderivative is available. It is exact in the sense that we can find the desired area to any prescribed accuracy you specify by summing sufficiently many terms in the series.
Objecting to such a situation is akin, IMO, to objecting to the fact that $\sqrt 2$ is not rational. Ok, it's not rational, but we can still compute $\sqrt 2$ exactly in the same sense that we computed the area above exactly. The point is to be able to compute the quantity of interest to a prescribed level of accuracy; I am indifferent as to the form the particular computation takes (apart from mere subjective personal preference).
A: We can, and do all the time! For example, 
The Gamma function
$$
\Gamma (z) = \int_0^{\infty} t^{z-1}\mathrm{e}^{-t} \ \mathrm{d}t.
$$
The Beta function 
$$
\mathrm{B}(z,y) = \int_0^1 t^{z-1}(1-t)^{y-1}\,\mathrm{d}t.
$$
The Exponential integral function
$$
\mathrm{E}_1(z) = \int_z^\infty \frac{e^{-t}}{t}\,  \mathrm{d}t.
$$
The Error function
$$
\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt.$$
The Elliptic integral of the second kind
$$
 E(\phi,k)=\int_0^{\phi} \sqrt{1-k^2\sin^2\theta} \ \mathrm{d}\theta. 
$$
The Logarithmic integral function
$$
 {\rm Li} (x) = \int_2^x \frac{\mathrm{d}t}{\ln t},
$$
and many, many more very important "special functions' are defined by definite integrals. If you go back even to elementary functions, you can define the logarithm via the following integral - 
The Logarithm
$$
\ln (t) = \int_1^t \frac{1}{x} \, dx.
$$
As to your first question, how do we find areas under these curves if we don't have an elementary antiderivative? Well, how do you find the area under the curve $1/t$ from $1$ to $5$? The above integral tells you the value is $\ln 5$, but what is that value, exactly? We can only approximate it, given the best methods we have! 
The same is true of all the above functions. At some special values they have exact values, given perhaps by integers, rational or irrational numbers, or a combination of common mathematical constants such as $\pi,\mathrm{e},\gamma,$ Catalan's constant, etc. (another interesting question is - why are these constants special enough to have names? Because they come up all the time! The same is true for the above functions) .
But for almost all values we must approximate the value of the function by computing the definite integral numerically (Trapezoid rule, Simpson's rule, more advanced techniques), or using some other representation of the integral such as an infinite sum etc. 
