How to Use the Definition of the Definite Integral to Evaluate .
The definition of the definite integral evaluates $\int_a^b f(x),dx$ as the limit of a Riemann sum, $\lim_{|\Delta|\to 0}\sum_{i=1}^{n} f(x_i^*),\Delta x_i$, where the interval $[a,b]$ is sliced into subintervals, each contributing a rectangle’s area. To use the definition of the definite integral to evaluate a given integral, you partition the interval, build the sum, simplify it with summation formulas, and take the limit as the partition gets finer. This direct method returns the same value as the Fundamental Theorem of Calculus, and it is the more instructive route when you want to see where the integral actually comes from.
Riemann Sum Definition
A Riemann sum is a finite sum that approximates the signed area under a curve by adding up rectangle areas. A partition $P$ of $[a,b]$ is a set of points $a = x_0 < x_1 < \dots < x_n = b$; the $k$-th subinterval is $[x_{k-1}, x_k]$ with width $\Delta x_k = x_k - x_{k-1}$. A sample point $x_k^$ is any point chosen inside that subinterval. The sum $\sum_{k=1}^{n} f(x_k^),\Delta x_k$ then adds one rectangle per subinterval, using the function value at the sample point as the rectangle’s height.
The definite integral is defined as the limit of these sums as the partition becomes infinitely fine — formally, as the norm of the partition, $|\Delta| = \max_k \Delta x_k$, goes to zero:
$$\int_a^b f(x),dx = \lim_{|\Delta|\to 0}\sum_{k=1}^{n} f(x_k^*),\Delta x_k$$
Two facts make this definition workable. First, when $f$ is continuous, the limit exists and gives the same number no matter how you choose the sample points. Second, area above the x-axis counts positively while area below counts negatively, which is why the integral is called signed area. That signed-area interpretation is the standard geometric picture used throughout calculus references such as Wikipedia’s list of definite integrals.
Regular Partitions Simplify Everything
For the limit computation you almost always use an equal-width partition, where every subinterval has the same width $\Delta x = \frac{b-a}{n}$ and $x_k = a + k\Delta x$. This choice is allowed because the definition requires the limit as the norm shrinks — and for a regular partition, $|\Delta| = \Delta x = \frac{b-a}{n} \to 0$ exactly when $n \to \infty$.
The most convenient sample points are the right endpoints, $x_k^* = x_k = a + k\Delta x$, though left endpoints or midpoints work just as well and produce the same final value for a continuous integrand.
Setting Up Riemann Sums
To build the sum for a particular function, follow the same four moves every time:
- Find $\Delta x = \frac{b-a}{n}$.
- Write the grid points $x_k = a + k\Delta x$.
- Evaluate $f(x_k)$ and multiply by $\Delta x$ to get the $k$-th rectangle’s area.
- Sum from $k = 1$ to $n$.
Suppose $f(x) = x^2$ on $[0, 2]$. Then $\Delta x = \frac{2}{n}$ and $x_k = \frac{2k}{n}$. Since $f(x_k) = \left(\frac{2k}{n}\right)^2 = \frac{4k^2}{n^2}$, the sum is
$$S_n = \sum_{k=1}^{n} \frac{4k^2}{n^2} \cdot \frac{2}{n} = \frac{8}{n^3}\sum_{k=1}^{n} k^2.$$
The trick that makes this tractable is pulling every factor that does not depend on $k$ outside the summation sign. What remains is a sum of pure powers of $k$, which has a closed form.
The Power-Sum Formulas You Need
Three identities cover nearly every textbook problem of this type, and MIT OpenCourseWare’s calculus materials rely on them routinely (MIT OpenCourseWare):
| Sum | Closed form |
|---|---|
| $\sum_{k=1}^{n} 1$ | $n$ |
| $\sum_{k=1}^{n} k$ | $\frac{n(n+1)}{2}$ |
| $\sum_{k=1}^{n} k^2$ | $\frac{n(n+1)(2n+1)}{6}$ |
Notice that the first row simply counts terms, so $\sum_{k=1}^n c = cn$ for any constant $c$. Higher powers such as $\sum k^3$ exist too, but these three handle linear, quadratic, and constant integrands — which is the whole scope of a typical definition-based exercise.
Evaluating the Limit
Once the sum is written as a rational expression in $n$, evaluate the limit as $n \to \infty$. For the $f(x) = x^2$ example, substitute the closed form:
$$S_n = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{8}{6} \cdot \frac{(n+1)(2n+1)}{n^2} = \frac{4}{3}\cdot\frac{2n^2 + 3n + 1}{n^2}.$$
Expanding and dividing term by term gives $\frac{4}{3}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)$. Every term with an $n$ in the denominator vanishes as $n \to \infty$, leaving $\frac{4}{3}\cdot 2 = \frac{8}{3}$.
So $\int_0^2 x^2,dx = \dfrac{8}{3}$, which matches the familiar antiderivative check $\left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$.
Worked Example
Compute $\int_0^3 (2x + 1),dx$ directly from the definition, using right endpoints.
Step 1 — Partition. With $n$ equal subintervals, $\Delta x = \frac{3-0}{n} = \frac{3}{n}$ and $x_k = \frac{3k}{n}$.
Step 2 — Sample and build the sum. The right endpoint value is $f(x_k) = 2\cdot\frac{3k}{n} + 1 = \frac{6k}{n} + 1$, so each rectangle has area $\left(\frac{6k}{n} + 1\right)\frac{3}{n}$. Summing over $k$ splits into two pieces:
$$S_n = \frac{18}{n^2}\sum_{k=1}^{n} k + \frac{3}{n}\sum_{k=1}^{n} 1.$$
Step 3 — Apply the formulas. Using $\sum k = \frac{n(n+1)}{2}$ and $\sum 1 = n$, this becomes $\frac{18}{n^2}\cdot\frac{n(n+1)}{2} + 3 = 9\cdot\frac{n+1}{n} + 3$.
Step 4 — Take the limit. As $n \to \infty$, the fraction $\frac{n+1}{n} \to 1$, so the sum tends to $9 + 3 = 12$.
Indeed, $\int_0^3 (2x+1),dx = \left[x^2 + x\right]_0^3 = 9 + 3 = 12$, confirming the result. A useful sanity check: the region is a trapezoid with parallel sides of height $f(0) = 1$ and $f(3) = 7$ and width $3$, giving area $\frac{1+7}{2}\cdot 3 = 12$. Khan Academy’s practice sets follow this identical four-step rhythm for polynomial integrands (Khan Academy).
When the algebra gets heavy, remember that the definition and the Fundamental Theorem must agree for every continuous function — so the antiderivative is always a valid cross-check on your limit computation.
Common Pitfalls When Taking the Limit
Most errors in definition-based integration happen before the limit is even reached. Watch for these.
- Forgetting to substitute the closed form. The expression $\sum_{k=1}^n k^2$ is not $\frac{n^3}{3}$; using the exact $\frac{n(n+1)(2n+1)}{6}$ first, then expanding, keeps the limit exact. Only discard lower-order terms after dividing by the highest power of $n$.
- Distributing $f$ incorrectly. For $f(x) = x^2 + 3x$, write $f(x_k) = \frac{4k^2}{n^2} + \frac{6k}{n}$ as separate sums before applying formulas — square the whole bracket $(x_k)^2$ only, not the $3x$ term.
- Mis-setting the grid points. With left endpoints the k-th sample is $a + (k-1)\Delta x$, not $a + k\Delta x$; an off-by-one shift changes the final answer.
- Assuming a nonpolynomial integrand is impossible. Functions like $f(x) = \sin x$ or $f(x) = e^x$ have limit-of-sum representations too, but they require identities (sum-of-sines, geometric series) rather than power-sum formulas. The structural method is identical: build the sum, simplify, let $n \to \infty$.
Working through several of these by hand — polynomials first, then constants, then negative regions — trains the same muscle the Fundamental Theorem later spares you from using, which is exactly the pedagogical payoff of the limit definition.
Key Takeaways
- The definite integral is defined as $\lim_{|\Delta|\to 0}\sum_{k=1}^{n} f(x_k^*),\Delta x_k$, the limit of rectangle areas as the finest subinterval shrinks to zero.
- A regular partition sets $\Delta x = \frac{b-a}{n}$ and $x_k = a + k\Delta x$; right endpoints are the usual choice of sample point.
- Convert the sum into a rational expression using the power-sum formulas for $\sum 1$, $\sum k$, and $\sum k^2$, then let $n \to \infty$.
- Terms with $n$ in the denominator drop out in the limit; what survives is the exact value of the integral.
- Because continuous functions give the same limit regardless of sample points, you can always verify the answer against the antiderivative method.
- For integrals over infinite intervals, the definition is extended with an additional limiting step, giving an improper integral — see the general treatment at Wikipedia.
