Area Under y = x^2 on [0,1] via First-Principles (Riemann-Sum) Integration

Problem Statement

Goal: Compute the exact area of the region
- Under the parabola y = x^2
- Above the x-axis
- Bounded horizontally by the closed interval [0,1].
Key observation: The region is not a simple geometric shape (e.g.
triangle, rectangle, trapezoid), so classical area formulas do not apply; we must integrate.

Foundational formula used:
- \displaystyle \int{a}^{b} f(x)\,dx = \lim{n\to\infty} \sum{i=1}^{n} f(xi)\,\Delta x
Interpretation:
- Subdivide [a,b] into n equal subintervals.
- On each subinterval, erect a rectangle whose height is determined by a sample point x_i.
- Add the rectangle areas f(x_i)\,\Delta x and let n \to \infty to make width \Delta x go to 0.
Geometric connection: More rectangles ⇒ better approximation⇒ limit gives exact area.

Function: f(x)=x^2 (from “under the parabola”).
Interval endpoints:
- Left a=0 (given),
- Right b=1 (given).
Width of each subinterval/rectangle (common width):
\displaystyle \Delta x = \frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}.
Choice of sample point xi (right-hand endpoints referenced in the transcript): \displaystyle xi = a + i\,\Delta x = 0 + i\left(\frac{1}{n}\right)=\frac{i}{n}.

Substitute each piece into the first-principles formula:
\displaystyle \int{0}^{1} x^2\,dx = \lim{n\to\infty} \sum{i=1}^{n} f(xi)\,\Delta x \;=\; \lim{n\to\infty} \sum{i=1}^{n} \bigl(\,(x_i)^2\bigr)\,\Delta x.
Replace xi and \Delta x with their explicit expressions: xi = \frac{i}{n}, \quad \Delta x = \frac{1}{n}.
Obtain:
\displaystyle \lim{n\to\infty} \sum{i=1}^{n} \Bigl(\frac{i}{n}\Bigr)^2 \frac{1}{n}
= \lim{n\to\infty} \sum{i=1}^{n} \frac{i^2}{n^3}.

Standard result (memorize!):
\displaystyle \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}
(Same as transcript’s algebraically equivalent \tfrac{n(2n^2+3n+1)}{6}).

Plug into the working expression:
\displaystyle \int{0}^{1} x^2\,dx = \lim{n\to\infty} \frac{1}{n^3} \cdot \frac{n(2n^2+3n+1)}{6}.
Multiply numerators / denominators:
= \lim_{n\to\infty} \frac{2n^2 + 3n + 1}{6n^2}.
Degree analysis (highest power n^2 top & bottom):
- Leading coefficients: numerator 2, denominator 6.
- Therefore, as n \to \infty,
  \displaystyle \lim_{n\to\infty} \frac{2n^2 + 3n +1}{6n^2} = \frac{2}{6} = \frac{1}{3}.

Exact area under y=x^2 from 0 to 1:
\boxed{\displaystyle \frac{1}{3}}.
Checks/intuition:
- For y = x^2 over [0,1], the maximum height is 1, so the area must be < 1 \times 1 = 1, making 1/3 plausible.

Confirms result obtainable by the Fundamental Theorem of Calculus:
\int{0}^{1} x^2\,dx = \Bigl[\frac{x^3}{3}\Bigr]{0}^{1} = \frac{1}{3}-0 = \frac{1}{3}.
First-principles method matches more advanced rule.
Demonstrates the bridge between limit-based (Riemann) definition and antiderivative technique.
Highlights the usefulness of summation formulas in calculus.

Ethically, showing the derivation from first principles avoids reliance on “black-box” rules and promotes deeper comprehension.
Pedagogical payoff: reinforces
- Summation manipulation skills,
- Comfort with limits,
- Understanding why integrals equal areas.

Always identify f(x), a, b first; the rest follows mechanically.
Keep track of constants vs. index-dependent terms inside sums.
Memorize or have quick access to common sum formulas (e.g., \sum i, \sum i^2, \sum i^3).
In limit evaluation of rational functions, compare highest powers to determine limit quickly.
Remember alternative sample points (left endpoint, midpoints) produce same limit for continuous functions but can lead to different intermediate algebra.