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Definition of the Definite Integral

The definition of the definite integral is the foundation of integration theory; via Riemann sums it gives a rigorous meaning to “accumulation.”

Riemann sum definition

Riemann sums were introduced by Bernhard Riemann (1826–1866), laying rigorous groundwork for definite integrals.

Let $f(x)$ be defined on $[a, b]$ , partition it into $n$ subintervals:

$a = x_0 < x_1 < \cdots < x_n = b$

Pick any point $\xi_i$ in each $[x_{i-1}, x_i]$ , and form

$S_n = \sum_{i=1}^n f(\xi_i) \Delta x_i$

where $\Delta x_i = x_i - x_{i-1}$ . If $\lim_{\max\{\Delta x_i\}\to 0} S_n$ exists, it is the definite integral of $f(x)$ on $[a,b]$ :

$\int_a^b f(x) \, dx = \lim_{\max\{\Delta x_i\} \to 0} \sum_{i=1}^n f(\xi_i) \Delta x_i$

$\int$ (integral sign): from the elongated “S” of Latin summa, introduced by Leibniz (1646–1716).
Reading: $\int_a^b f(x) \, dx$ — “integrate $f(x)$ from $a$ to $b$ .”

Geometric meaning

$\int_a^b f(x) \, dx$ is the (signed) area of the curvilinear trapezoid bounded by $y=f(x)$ , $x=a$ , $x=b$ , and the $x$ -axis.

If $f(x)\ge 0$ , it’s the area.
If $f(x)\lt0$ , it’s the negative of the area.
If $f(x)$ changes sign, it’s the algebraic sum of signed areas.

Existence

Theorem

If $f(x)$ is continuous on $[a,b]$ , then $f(x)$ is integrable on $[a,b]$ .

Corollaries:

Continuous functions on closed intervals are integrable.
Bounded functions with finitely many discontinuities are integrable.
Monotone functions on closed intervals are integrable.

Practice Problems

Exercise 1

Using the Riemann-sum definition, approximate $\int_0^1 x \, dx$ with $n=4$ midpoints.

Reference Answer

riemann sumdefinite integral

Approach: Use the Riemann-sum definition with $n=4$ equal subintervals of $[0,1]$ . Each has length $\Delta x = \tfrac14$ , and choosing the midpoint of each subinterval gives the sample points $\xi_i$ .

Steps:

Split $[0,1]$ into four equal parts, so $\Delta x = \tfrac14$ .
The midpoints are $\xi_1=\tfrac18$ , $\xi_2=\tfrac38$ , $\xi_3=\tfrac58$ , $\xi_4=\tfrac78$ .
Build the sum: $S_4=\tfrac14\left(\tfrac18+\tfrac38+\tfrac58+\tfrac78\right)=\tfrac14\cdot\tfrac{16}{8}=\tfrac12$ .
Refining the mesh keeps the same limit, so $\int_0^1 x\,dx=\tfrac12$ .

Answer: $\tfrac12$ .

Exercise 2

Explain the geometric meaning of $\int_{-1}^1 x \, dx$ .

Reference Answer

geometric meaningdefinite integral

Approach: Interpret $\int_{-1}^1 x\,dx$ as a signed area. The integrand $f(x)=x$ is negative on $[-1,0]$ and positive on $[0,1]$ , so the two triangular regions contribute with opposite signs.

Steps:

On $[-1,0]$ the graph lies below the $x$ -axis, giving a signed area of $-\tfrac12$ .
On $[0,1]$ the graph lies above the $x$ -axis, giving a signed area of $+\tfrac12$ .
Add the signed areas: $-\tfrac12 + \tfrac12 = 0$ .

Answer: The algebraic sum of the signed areas is $0$ .

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\xi$	Greek letter	Xi（ksee）	An arbitrary point inside a subinterval
$\Delta x_i$	symbol	Delta x_i	The length of the $i$ -th subinterval
$\sum$	Greek letter	Sigma	The summation sign
$\lim$	mathematical symbol	limit	The limit
$\int$	mathematical symbol	integral	Definite integral sign

Periodicity of Integrals

Newton–Leibniz Formula

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