The integral of a continuous function is the limit of any Riemann sums whose pieces shrink, because the sums are squeezed between lower and upper sums whose gap goes to $0$ (the integral is well defined). Read backward, this turns certain limits of sums into integrals you can evaluate with the fundamental theorem.
🔑 Key idea
If $f$ is continuous on $[0, 1]$, then $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right) = \int_0^1 f(x)\,dx.$$ To use it, rewrite each term as $f\left(\frac{k}{n}\right)\cdot\frac{1}{n}$: the sample point is $\frac{k}{n}$ and the width is $\frac{1}{n}$.
💡 Why it works
The points $\frac{k}{n}$ cut $[0, 1]$ into $n$ pieces of width $\frac{1}{n}$, and $f\left(\frac{k}{n}\right)$ is the value at the right end of the $k$th piece. So the sum is a Riemann sum, and it lies between $L(f, P_n)$ and $U(f, P_n)$. Since $f$ is continuous, it is integrable, so $U - L \to 0$ and both tend to $\int_0^1 f$. The sum is squeezed to the same value. The same idea works on other intervals: $\sum_{k=1}^{2n} \frac{1}{n} f\left(\frac{k}{n}\right) \to \int_0^2 f$.
Right-endpoint sums for $\frac{1}{1 + x^{2}}$ on $[0, 1]$ approach $\frac{\pi}{4}$: they are exactly the sums $\sum \frac{n}{n^{2} + k^{2}}$.
The recognition step is the whole skill. Look for a factor $\frac{1}{n}$ (or something that becomes one after dividing top and bottom by a power of $n$), and for $k$ appearing only through $\frac{k}{n}$. Then the function is whatever is applied to $\frac{k}{n}$, and the interval runs from where $\frac{k}{n}$ starts to where it ends. A constant factor in front of the sum just multiplies the integral.
Worked examples
✏️ Worked example: A sum that becomes $\frac{\pi}{4}$
Answer: $\int_0^1 x^{3}\,dx = \frac{1}{4}$ (compare $\sum_{k=1}^{n} k^{3} = \frac{n^{2}(n + 1)^{2}}{4}$, which gives the same limit).
⚠️ Common mistake
Losing the width or the interval. Every term must be written as $f(\text{point})\cdot\frac{1}{n}$; a sum like $\sum \frac{k}{n^{2}}$ hides the $\frac{1}{n}$ inside $\frac{k}{n}\cdot\frac{1}{n}$. And count the terms: $k$ from $1$ to $2n$ with width $\frac{1}{n}$ covers $[0, 2]$, not $[0, 1]$, so $\sum_{k=1}^{2n} \frac{1}{n + k} \to \ln 3$.
These limits are a two-way street. Numerical methods use them forward: a Riemann sum with large $n$ approximates an integral. Analysis uses them backward: a limit of sums that looks hopeless by algebra is recognized as an integral and evaluated exactly.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.