MA 16500 Notes | Hanh Vo

Goal: Understand the definition of the definite integral, its interpretation as net area, and its connection to antiderivatives.

1. Introduction to Definite Integrals

Definition (Definite Integral): Let \( f(x) \) be a function defined on the interval \([a,b]\). The definite integral of \(f\) from \(a\) to \(b\) is defined as

\[ \int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*) \,\Delta x \]

where \( \Delta x = \frac{b-a}{n} \) and \( x_k^* \) is any point in the \(k\)-th subinterval.

Remark (Geometric Interpretation):

The definite integral represents the net area between the graph of \(f(x)\) and the \(x\)-axis.
Areas above the \(x\)-axis are positive.
Areas below the \(x\)-axis are negative.

\[ \int_a^b f(x)\,dx = (\text{area above the } x\text{-axis}) - (\text{area below the } x\text{-axis}) \]

Remark (Connection to Indefinite Integrals):

If \(F(x)\) is an antiderivative of \(f(x)\), then

\[ \int_a^b f(x)\,dx = F(b) - F(a) \]

This result is known as the Fundamental Theorem of Calculus (Part 2).

Properties of Definite Integrals

Property	Statement	Remarks / Explanation
Linearity	\(\displaystyle \int_a^b [c f(x) + d g(x)]\,dx = c \int_a^b f(x)\,dx + d \int_a^b g(x)\,dx\)	Holds for constants \(c, d \in \mathbb{R}\).
Additivity over intervals	\(\displaystyle \int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx\)	Can split the integral at any point \(c\) in \([a,b]\).
Reversing limits	\(\displaystyle \int_a^b f(x)\,dx = - \int_b^a f(x)\,dx\)	Swapping the bounds reverses the sign of the integral.
Positivity	If \(f(x) \ge 0\) for all \(x \in [a,b]\), then \(\displaystyle \int_a^b f(x)\,dx \ge 0\)	Geometrically: area above x-axis is non-negative.
Comparison	If \(f(x) \ge g(x)\) for all \(x \in [a,b]\), then \(\displaystyle \int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx\)	Useful for inequalities and estimating integrals.

2. Examples

Example 2.1: Riemann Sum and Geometric Interpretation

Consider the integral \(\displaystyle \int_0^1 \sqrt{1 - x^2}\, dx\).

Write down the formula for approximating the integral as a Riemann sum by dividing the interval \([0,1]\) into \(n\) equal subintervals and using the right endpoints.
Compute the integral using the geometric interpretation.

Solution:

1. Riemann Sum Approximation:

Divide the interval \([0,1]\) into \(n\) equal parts:

\[ \Delta x = \frac{1-0}{n} = \frac{1}{n} \]

Right endpoints:

\[ x_k = \frac{k}{n}, \quad k = 1,2,\dots,n \]

Riemann sum:

\[ \sum_{k=1}^{n} \sqrt{1 - \left(\frac{k}{n}\right)^2} \cdot \frac{1}{n} \]

Integral as a limit:

\[ \int_0^1 \sqrt{1 - x^2}\, dx = \lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{1 - \left(\frac{k}{n}\right)^2} \cdot \frac{1}{n} \]

2. Geometric Interpretation:

The function \(y = \sqrt{1 - x^2}\) represents the upper semicircle of radius \(1\). The integral \(\int_0^1 \sqrt{1 - x^2}\, dx\) gives the area under the curve from \(0\) to \(1\).

The area of a quarter circle of radius \(1\) is

\[ \text{Area} = \frac{1}{4} \pi (1)^2 = \frac{\pi}{4} \]

Thus, using the geometric interpretation:

\[ \int_0^1 \sqrt{1 - x^2}\, dx = \frac{\pi}{4} \]

Example 2.2: Riemann Sum Approximation

Compute the integral \(\displaystyle \int_0^2 (x^3 + 1)\, dx\) using a Riemann sum by dividing the interval \([0,2]\) into \(n\) equal subintervals and using right endpoints.

Solution:

Step 1: Divide the interval \([0,2]\) into \(n\) equal subintervals:

\[ \Delta x = \frac{2-0}{n} = \frac{2}{n} \]

Step 2: Right endpoints:

\[ x_k = 0 + k \Delta x = \frac{2k}{n}, \quad k=1,2,\dots,n \]

Step 3: Riemann sum:

\[ \begin{aligned} \sum_{k=1}^{n} \left[ \left(\frac{2k}{n}\right)^3 + 1 \right] \cdot \frac{2}{n} &= \sum_{k=1}^{n} \frac{2}{n} \left( \frac{8 k^3}{n^3} + 1 \right) \\[1mm] &= \sum_{k=1}^{n} \frac{16 k^3}{n^4} + \sum_{k=1}^{n} \frac{2}{n} \\[1mm] &= \frac{16}{n^4} \sum_{k=1}^{n} k^3 + \frac{2}{n} \sum_{k=1}^{n} 1 \\ &= \frac{16}{n^4} \sum_{k=1}^{n} k^3 + \frac{2}{n} \cdot n \\ &= \frac{16}{n^4} \sum_{k=1}^{n} k^3 + 2 \end{aligned} \]

Step 4: Use the formula \(\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}\):

\[ \frac{16}{n^4} \cdot \frac{n^2(n+1)^2}{4} + 2 = \frac{4(n+1)^2}{n^2} + 2 \]

Step 5: Take the limit as \(n \to \infty\):

\[ \int_0^2 (x^3 + 1)\, dx = \lim_{n\to\infty} \left( \frac{4(n+1)^2}{n^2} + 2 \right) = 4 + 2 = 6 \]

Thus, the integral evaluates to:

\[ \int_0^2 (x^3 + 1)\, dx = 6 \]