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Section4.3Properties of Definite Integrals

Subsection4.3.1Overview

Motivated by the properties of total accumulated change and of area, the definite integral inherits several significant properties. These properties are stated as theorems. We will be interested in applying the results of the theorems. However, to prove these properties is outside the scope of this text. We essentially think of these properties as the axioms of definite integrals, basic properties which must always be true.

Subsection4.3.2Integrability

The process of computing a definite integral is well-defined for simple functions. However, other functions have potential issues. One of the most significant developments of modern mathematics was developing an understanding of when functions can be integrated or not, even going so far as to create new definitions for integrals for different circumstances. For our purposes, we will focus on the more elementary Riemann integral which is based on Riemann sums.

Definition4.3.1Riemann Integrability

A function \(f : [a,b] \to \mathbb{R}\) is Riemann integrable (or, more simply, integrable) on \([a,b]\) if \(\displaystyle \int_{a}^{b} f(x) \, dx\) is defined. For our purposes, we will interpret this as saying that for every scheme of constructing a Riemann sum with partition of size \(n\) (i.e., any choices of evaluation points \(x_k^* \in [x_{k-1},x_k]\)), the limit of the Riemann sum is guaranteed to equal the same value: \begin{equation*} \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*) \Delta x = \int_{a}^{b} f(x) \, dx. \end{equation*}

The scope of mathematics for this text is not concerned with determining which functions are or are not integrable, with the exception of pointing out that continuous functions are integrable.

We introduced the idea of definite integrals using simple functions. Simple functions, which are piecewise constant, are clearly not continuous because of the jumps at points in the partition. Integrability must be more general than continuous functions. For our purposes, we will also include functions that are piecewise continuous, allowing a finite number of either removable discontinuities or jump discontinuities. This is based on the idea that including or excluding the endpoints of an interval does not change integrability.

Proof

While talking about conditions where a function is integrable, we should also include some discussion about where a function is not integrable. The most common issue that prevents integrability is when a function is unbounded, such as at a vertical asymptote. Such a discontinuity is called an infinite discontinuity. Definite integrals generally do not behave well at such discontinuities and special techniques must be developed later in calculus to analyze how to address the discontinuity. For now, we will consider all such functions to be nonintegrable.

Subsection4.3.3Splitting and Linearity Properties

Splitting properties are motivated by considering adjacent intervals, say \([a,b]\) and \([b,c]\), and requiring that the definite integral on \([a,c]\) is the sum of the integrals over the two pieces. However, this would require that \(a \lt b \lt c\). While this is certainly true, the definite integral is defined in a way that the order does not matter. Recalling that the definite integral is motivated as the mathematical tool to compute the total change in a quantity as the accumulated change resulting from its rate of change, this result could be interpreted as saying, “The total change in \(Q\) as \(x\) goes from \(a\) to \(c\) is equal to the change in \(Q\) as \(x\) goes first from \(a\) to \(b\) plus the change as \(x\) then goes from \(b\) to \(c\).”

Similarly, if \(x\) does not change, then the dependent quantity \(Q\) should also not change, regardless of the function defining the rate of change. This is the motivation for the next theorem.

Combining these theorems, we obtain a reversal property of definite integrals. If we switch the order of the limits of integration, then the value of the definite integral must change sign.

Proof

It is important to remember that the interpretation of a definite integral as signed area only applied when the limits of integration go from left to right. When the limits of integration go from right to left, we compute the signed area but need to reverse the sign because of the direction of integration.

Example4.3.8

Suppose that we know \(\displaystyle \int_{0}^{4} f(x)\,dx = 6\) and \(\displaystyle \int_{3}^{4} f(x) \, dx = 10\). Find \(\displaystyle \int_{0}^{3} f(x) \, dx\).

Solution
Example4.3.9

Suppose that the graph below shows \(y=f(x)\). Use the graph to find \(\displaystyle \int_{0}^{-3} f(x) \, dx\).

<<SVG image is unavailable, or your browser cannot render it>>

Solution

Because a definite integral is defined in terms of summation and summation has linearity properties, the definite integral inherits a constant multiple rule and a sum rule.

We often want to apply these two properties immediately after one another. The combination of the constant multiple rule and sum rule is called linearity.

These rules allows us to take elementary functions and combine them using linear combinations (sums and constant multiples) to create more complicated functions and compute the definite integral as the corresponding linear combination of the integrals of the components. The next subsection introduces some of these elementary functions and their corresponding accumulations defining definite integrals.

Example4.3.13

Suppose that we know \(\displaystyle \int_{1}^{4} f(x) \, dx = 8\) and \(\displaystyle \int_{1}^{4} g(x) \, dx = -3\). Find the values of the following definite integrals.

  1. \(\displaystyle \int_{1}^{4} 2 f(x) \, dx\)

  2. \(\displaystyle \int_{1}^{4} [f(x)-2g(x)] \, dx\)

  3. \(\displaystyle \int_{1}^{4} [3+g(x)] \, dx\)

Solution

The final property of definite integrals relates to inequality relationships.

Subsection4.3.4Elementary Accumulations

We introduce a few elementary functions for which definite integrals can be computed. Each of the definite integrals have a lower limit of integration at \(x=0\). To compute definite integrals over other intervals, we use the splitting properties of definite integrals. You should notice that these definite integrals are closely related to elementary summation formulas.

Proof
Example4.3.16

Compute \(\displaystyle \int_0^4 (3x-4) dx\)

Solution
Example4.3.17

Compute \(\displaystyle \int_0^3 (2x+1)(x-3) dx\)

Solution

When the lower limit is not zero, we need to use the splitting property of definite integrals.

Example4.3.18

Compute \(\displaystyle \int_{2}^{5} 2x^2 \, dx\).

Solution