To compute the area under a curve. We need to take every point where the function is defined and sum up all the tiny line we obtain. This is $$\int_{a}^{b}f(x)dx$$
We are summing up tiny rectangles of height $f(x)$ and width $dx$. But because there are infintely many points in the interval $[a,b]$ and because a point has no width, we can't sum up all of them or we will obtain of result of $\infty$. That's why we sum up with a width of $dx$. But by doing that we are not really precise on the computation, this is not the smallest measurment possible, we have a roundness error right ?
The correct way (for me) of defining an integral should be :
$$\lim_{\text{dx}\rightarrow0}\int_{a}^{b}f(x)dx$$
Best Answer
I will walk you through a somewhat informal derivation of the Riemann integral.
Basically, you want to cut up an area into a bunch of tiny rectangles then add their areas.
So think of some function $f(x)$ which is continuous on $[a,b]$. Then let $$I=\int_a^bf(x)\mathrm dx$$ Our goal is to approximate this really well. To do so, we cut up $[a,b]$ into $n$ segments of equal length, and call each one of these segments $S_i=[\alpha_i,\beta_i]$ for integer $0\leq i\leq n$. Also we will want to define $\beta_i=\alpha_{i+1}$, and $\alpha_i<\beta_i$.
Thus $\text{length}(S_i)=\frac{b-a}n$.
We then make a rectangle with base $S_i$ and height $f(\beta_i)$. Call these rectangles $R_i$. So, it makes since that $$I\approx \sum_{i=0}^{n}\text{Area}(R_i)$$ When we recall that $\text{Area}(R_i)=\text{base}(R_i)\cdot\text{height}(R_i)=\frac{b-a}nf(\beta_i)$ (because $R_i$ is a rectangle), we have that $$I\approx \sum_{i=0}^n\frac{b-a}nf(\beta_i)=\frac{b-a}n\sum_{i=0}^nf(\beta_i)$$ Then we recall that by definition $\alpha_0=a$, $\beta_n=b$, and $\beta_i-\alpha_i=\frac{b-a}n$, we have that $$\beta_i=a+\frac{i}n(b-a)$$ So we have that $$I\approx \frac{b-a}n\sum_{i=0}^nf\bigg(a+\frac{i}n(b-a)\bigg)$$ And the larger we make $n$, the more rectangles there will be, thus the better our approximation of $I$ will be. Or more formally, $$I=\lim_{n\to\infty}\frac{b-a}n\sum_{i=0}^nf\bigg(a+\frac{i}n(b-a)\bigg)$$ and hence the definition of the definite Riemann integral: $$\int_a^bf(x)\mathrm dx=\lim_{n\to\infty}\frac{b-a}n\sum_{i=0}^nf\bigg(a+\frac{i}n(b-a)\bigg)$$ As you mentioned, the approximation of an integral with a sum is rarely (if ever) going to be perfect. This it to be expected. Which is why the $\lim_{n\to\infty}$ is there. The limit ensures the equality of the integral and the sum. Then integral is an exact value, and the sum approaches it.
I hope this helped.