Defining limit of integral using fractions

Question

I was trying to derive limit form of integral and I got the below. But I'm not sure what is wrong about it. I learned about it in class today and I'm having all the rectangles under a function summed since the width can be defined by $b/r$ and the height can be $f(bx/r)$

$$\int_{0}^{b} f(x) \,dx = \lim_{r \to \infty} \left(\sum_{k=0}^{r-1} \frac{b}{r}f\left(\frac{bk}{r}\right)\right)$$

Here I am setting $r$ to the number of rectangles (it approaches infinity), $f$ is the function and inside the summation it slowly sums each value until $b$ is reached.

I even tried running it using Python and it seems to work correctly for the basic functions I've tried

def calculate_limit(f, r):
    total = 0
    for x in range(r + 1):
        total += a/r * f(x*a / r)
    return total

def f(x):
    return x**2
    
r = 1000
a = 2

limit = calculate_limit(f, r)
print(limit)
```

Answer 1

(This is a community wiki answer! Please feel free to improve it to your liking.)

Nice self discovery!

Indeed there is no particularly wrong part of your discovery. The equality holds if $f$ is continuous (or even Riemann integrable) on the interval $[0, b]$:

\begin{align*} \int_{0}^{b} f(x) \, \mathrm{d}x &= \lim_{n\to\infty} \sum_{k=1}^{n} f\left(\frac{kb}{n}\right)\frac{b}{n} \tag{1}\\ &= \lim_{n\to\infty} \sum_{k=0}^{n-1} f\left(\frac{kb}{n}\right)\frac{b}{n} \tag{2} \end{align*}

Remarks:

• In $\text{(1)}$, the right endpoint of the subinterval is used to calculate the height of each rectangle, whereas in $\text{(2)}$, the left endpoint is used.

• Although the above equality is enough to establish a theoretical definition of a (rudimentary version of) definite integral, you may have already noticed that the approximation still deviates from the actual value even when the number $n$ of divisions is large. This opens up the question that eventually leads us to the area of numerical integration, where a much more efficient ways of approximating integrals has been studied.

---

I also brushed up your code for fun:

from typing import Callable


def calculate_limit(f: Callable[[float], float], n: int = 10, b: float = 1.0) -> float:
    sum: float = 0
    dx = b / n
    for k in range(n):
        sum += f(k * dx) * dx
    return sum


for n in [10, 100, 1000, 10000, 100000]:
    limit = calculate_limit(lambda x: x**2, n, b=2)
    print(
        f"Simluated value using {n} divisions\n    = {limit}\n"
    )  # Exact value is 8/3 = 2.666...