Riemann sums — left, right, midpoint; formal definition of definite integral

WHY do we need this?

A rectangle's area is trivial: width × height. A curve's area is not. The whole trick of integration is to trade one hard area for infinitely many easy areas. The "Riemann sum" is the finite approximation; the "definite integral" is its limit.

WHAT is a partition? (build the scaffolding)

WHAT is a Riemann sum? (general form)

The only freedom is where you sample the height. Three classic choices:

DERIVATION: the formal definite integral from scratch

WHY take a limit? Each $S_n$ is only an approximation. As we use more, thinner rectangles ($n\to\infty$, $\Delta x\to 0$) the staircase hugs the curve ever tighter. The exact area is the limiting value.

Step 1 — define the "mesh". The mesh (norm) of a partition is the widest strip: $\|P\| = \max_i \Delta x_i$. Why this step? Shrinking $\|P\|\to 0$ forces every strip to vanish, not just the average.

Step 2 — demand the limit exists independent of choices.

Why "every choice"? If left, right, and midpoint sums all converge to the same number, the answer can't depend on our arbitrary sampling — it's a genuine property of $f$, not of our method.

Step 3 — uniform-partition shortcut. For continuous $f$ a uniform partition suffices, and $\|P\|\to0 \iff n\to\infty$: $\boxed{\ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f\big(a+i\,\Delta x\big)\,\frac{b-a}{n}, \quad \Delta x=\frac{b-a}{n}.\ }$

The notation now decodes itself: $\int$ is a stretched "S" for Sum; $dx$ is the limit of $\Delta x$ (an infinitely thin width); $f(x)$ is the height.

Worked Example 1 — $\int_0^1 x^2\,dx$ from the definition

We compute it as a limit of right sums.

$\Delta x = \frac{1-0}{n} = \frac1n$, and $x_i = 0 + i\cdot\frac1n = \frac{i}{n}$. Why this step? Right rule samples the right edge $x_i$.

$R_n = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac1n = \frac{1}{n^3}\sum_{i=1}^n i^2.$ Why this step? Factor out the constants $\frac1{n^2}\cdot\frac1n=\frac1{n^3}$, leaving a pure $\sum i^2$.

Use $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$: $R_n = \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}.$ Why this step? Plug the closed form so we can take a clean limit.

$\int_0^1 x^2\,dx = \lim_{n\to\infty}\frac{2n^2+3n+1}{6n^2} = \frac{2}{6} = \frac13.$ Why this step? Divide top & bottom by $n^2$; the $\frac3n,\frac1{n^2}$ terms die. ✓ (matches $\frac{x^3}{3}\big|_0^1$).

Worked Example 2 — numerical L, R, M for $\int_1^3 \tfrac1x\,dx$, $n=2$

True value $=\ln 3 - \ln 1 = \ln 3 \approx 1.0986$. $\Delta x = \frac{3-1}{2}=1$.

Left ($x_0=1, x_1=2$): $L_2 = (f(1)+f(2))\cdot1 = (1 + 0.5) = 1.5$. Why? Sample left edges $1$ and $2$.

Right ($x_1=2, x_2=3$): $R_2 = (f(2)+f(3))\cdot1 = (0.5 + 0.\overline{3}) = 0.8\overline{3}$. Why? Sample right edges $2$ and $3$.

Midpoint ($1.5, 2.5$): $M_2 = (f(1.5)+f(2.5))\cdot1 = (0.\overline{6}+0.4) = 1.0\overline{6}$. Why? Sample centres of each strip.

Notice $\frac1x$ is decreasing, so $L>$ truth $>R$, and $M\approx1.067$ is closest to $1.0986$. Exactly the prediction. ✓

Worked Example 3 — symbolic midpoint for $\int_0^2 (3x+1)\,dx$

Linear function ⇒ midpoint is exact for any $n$. Take $n=1$: $\Delta x=2$, midpoint $=1$. $M_1 = f(1)\cdot 2 = (3\cdot1+1)\cdot2 = 8.$ True: $\frac{3x^2}{2}+x\big|_0^2 = 6+2 = 8$. ✓ Why exact? On a line the triangle cut off above equals the triangle added below the midpoint height — perfect cancellation.

Common Mistakes (Steel-manned)

Active Recall

Connections

• Fundamental Theorem of Calculus — turns this limit into antiderivative evaluation $F(b)-F(a)$.
• Trapezoidal Rule — average of left & right sums.
• Simpson's Rule — weighted blend (midpoint + trapezoid) for higher accuracy.
• Summation formulas — $\sum i,\ \sum i^2,\ \sum i^3$ needed for limit evaluation.
• Area under a curve and Signed area — what the integral measures.
• Continuity and Integrability — sufficient conditions for the limit to exist.
• Limits of sequences — the $n\to\infty$ machinery.

Concept Map

Hinglish (regional understanding)