Riemann Sum Convergence: Watching n → ∞

The whole point of the integral is to take the limit of Riemann sums as the number of rectangles goes to infinity. This visualization makes the limit feel real: as you crank up $n$, the rectangles fill in the curve perfectly and the approximation converges to the exact value $64/3$.

The Limit Definition

$\int_{a}^{b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x$

where $\Delta x = (b - a)/n$ and $x_i^{*}$ is any sample point in the $i$-th subinterval. As $n \to \infty$ the partition becomes so fine that the sum converges to a single number — the definite integral.

Step-by-Step Solution

Given: $f(x) = x^{2}$, $a = 0$, $b = 4$. Find: how the right-endpoint Riemann sum behaves as $n$ grows, and confirm the limit.

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Step 1 — Use the closed-form for the right Riemann sum.

For the partition $x_i = i \cdot \dfrac{4}{n}$, the right Riemann sum is:

$R_n = \sum_{i=1}^{n} \left(\dfrac{4i}{n}\right)^{2} \cdot \dfrac{4}{n} = \dfrac{64}{n^{3}}\sum_{i=1}^{n} i^{2}$

Step 2 — Apply the famous formula $\sum_{i=1}^{n} i^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$.

$R_n = \dfrac{64}{n^{3}} \cdot \dfrac{n(n + 1)(2n + 1)}{6} = \dfrac{64\,(n + 1)(2n + 1)}{6 n^{2}}$

Step 3 — Tabulate $R_n$ for several $n$.

• $n = 4$: $R_4 = \dfrac{64 \cdot 5 \cdot 9}{6 \cdot 16} = \dfrac{2880}{96} = 30.000$
• $n = 8$: $R_8 = \dfrac{64 \cdot 9 \cdot 17}{6 \cdot 64} = \dfrac{9792}{384} = 25.500$
• $n = 16$: $R_{16} = \dfrac{64 \cdot 17 \cdot 33}{6 \cdot 256} \approx 23.375$
• $n = 32$: $R_{32} \approx 22.344$
• $n = 100$: $R_{100} \approx 21.547$
• $n = 1000$: $R_{1000} \approx 21.355$
• $n = 10000$: $R_{10000} \approx 21.336$

The sums are converging to $64/3 \approx 21.333$.

Step 4 — Take the formal limit.

$\lim_{n \to \infty} R_n = \lim_{n \to \infty} \dfrac{64\,(n + 1)(2n + 1)}{6 n^{2}} = \dfrac{64 \cdot 2}{6} = \dfrac{128}{6} = \dfrac{64}{3}$

(In the limit, $(n + 1)/n \to 1$ and $(2n + 1)/n \to 2$, so the whole expression goes to $64 \cdot 1 \cdot 2 / 6 = 64/3$.)

Step 5 — Verify using the antiderivative.

By the Fundamental Theorem of Calculus:

$\int_{0}^{4} x^{2}\,dx = \dfrac{x^{3}}{3}\bigg|_{0}^{4} = \dfrac{64}{3} - 0 = \dfrac{64}{3} \approx 21.333$

Both methods agree. ✓

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Answer: The Riemann sums for $\int_{0}^{4} x^{2}\,dx$ converge to $\dfrac{64}{3} \approx 21.333$. The convergence rate for the right-endpoint sum is $O(1/n)$ — error halves when you double $n$. The midpoint sum converges at $O(1/n^{2})$ — much faster.

Try It

• Slide n from 4 up to 100 — watch the error counter in the HUD shrink as the rectangles fill the area more snugly.
• The bottom plot tracks the approximation against $n$ — it homes in on the green dashed line at $64/3$ as $n$ grows.
• At $n = 100$, the sum is within ~1% of the exact value. At $n = 1000$ it'd be within 0.1%.