Lecture #22: Riemann Sums & Definite Integrals

MATH 145 – Calculus I

Torin Quinlivan

Knox College

November 6, 2025

Riemann Sums

Consider a function \(f(x)\) defined on an interval \(a \le t \le b\), and, given a whole number \(n\), divide the interval into \(n\) subintervals of length \(\Delta t = \dfrac{(b-a)}{n}\) with endpoints \(t_0, t_1, t_2, \dots, , t_n\).

The left (regular) Riemann sum is:

\[ f(t_0) \Delta t + f(t_1) \Delta t + \dots + f(t_{n-1}) \Delta t = \sum_{i=0}^{n-1} f(t_i) \Delta t. \]

Riemann Sums

The right (regular) Riemann sum is:

\[ f(t_1) \Delta t + f(t_2) \Delta t + \dots + f(t_n) \Delta t = \sum_{i=1}^{n} f(t_i) \Delta t. \]

Definite Integral

Given a continuous function \(f(x)\) on an interval \(a \le t \le b\), the definite integral of \(f(x)\) over \([a,b]\) is the limit:

\[ \int_a^b f(t)\,dt = \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i) \Delta t, \]

or:

\[ \int_a^b f(t)\,dt = \lim\limits_{n \to \infty} \sum_{i=0}^{n-1} f(t_i) \Delta t, \]

provided these limits exist. \(f(x)\) is called the integrand and \(a\) and \(b\) are called the limits of integration.

Example: Definite Integral

Example: Consider the function \(f(x) = x^2 + 1\) over the interval \([0, 2]\).

For a given \(n\), let \(\Delta t = \dfrac{2}{n}\) and \(t_i = \dfrac{2i}{n}\), so that:

\[ \int_0^2 x^2 + 1 \, dx = \lim\limits_{n \to \infty} \left(\left(\frac{2i}{n}\right)^2 + 1 \right) \times \left(\frac{2}{n}\right). \]

This sum can be thought of as a sum of areas of rectangles the graph of \(f\left(x\right)\).

Example: \(n=5\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{4} \left(\left(\frac{2i}{5}\right)^2 + 1\right) \times \left(\frac{2}{5}\right) = 4.384 \]

Example: \(n=10\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{9} \left(\left(\frac{2i}{10}\right)^2 + 1\right) \times \left(\frac{2}{10}\right) = 4.448 \]

Example: \(n=20\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{19} \left(\left(\frac{2i}{20}\right)^2 + 1\right) \times \left(\frac{2}{20}\right) = 4.571 \]

Example: \(n=50\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{49} \left(\left(\frac{2i}{50}\right)^2 + 1\right) \times \left(\frac{2}{50}\right) = 4.627264 \]

Example: \(n=100\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{99} \left(\left(\frac{2i}{100}\right)^2 + 1\right) \times \left(\frac{2}{100}\right) = 4.646808 \]

Example: \(n=1000\)

\[ \int_0^2 x^2 + 1 \, dx = \sum_{i=0}^{999} \left(\left(\frac{2i}{1000}\right)^2 + 1\right) \times \left(\frac{2}{1000}\right) = 4.664668 \]

Example: \(n \to \infty\)

As \(n \to \infty\), this sum becomes the area under the graph of \(f(x)\) over \([a,b]\).

\[ \int_0^2 x^2 + 1 \, dx = \lim\limits_{i \to \infty} \sum_{i=0}^{n-1} \left(\left(\frac{2i}{n}\right)^2 + 1\right) \times \left(\frac{2}{n}\right) = \frac{14}{3} \]

Estimates of \(\displaystyle\int_a^b f\left(x\right)\,dx\) can be obtained with left sums, right sums, sums from midpoint values of \(f(x)\), or even more general types of sums.

In simple cases, the values of \(\displaystyle\int_a^b f(x)\,dx\) can be found simply by finding an area under a graph.

Example: Definite Integral

Example: Evaluate \(\displaystyle\int_0^2 -2x + 1\,dx\).

Note: \(\displaystyle\int_a^b f(x)\,dx\) is the signed area – that is, the area between the graph of \(f(x)\) and the \(x\)-axis.

Velocity and Distance

Recall the definition of rate of change: \(r = \dfrac{d}{t}\).

This implies that \(d = rt\).

Suppose \(v(t)\) is the velocity of an object of an object at time \(t\) and consider the distance traveled over the time interval \(\Delta t\).

Velocity and Distance

\[ \begin{aligned} \int_a^b v(t) \, dt &= \lim\limits_{n \to \infty} \sum_{i = 1}^n v(t_i) \Delta t \\ &= \textrm{distance traveled (displacement)} \\ &= \textrm{area under the graph of } v(t) \end{aligned} \]

Example: Velocity and Distance

The velocity of an object moving along an axis is described by the picture below. Find the displacement of the object from time \(t=0\) to time \(t=4\).