Properties of the Definite Integral

Learning Outcomes

Use geometry and the properties of definite integrals to evaluate them

The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.

Properties of the Definite Integral

$\displaystyle\int_a^a f(x) dx = 0$

If the limits of integration are the same, the integral is just a line and contains no area.
$\displaystyle\int_b^a f(x) dx = −\displaystyle\int_a^b f(x) dx$

If the limits are reversed, then place a negative sign in front of the integral.
$\displaystyle\int_a^b [f(x)+g(x)] dx = \displaystyle\int_a^b f(x) dx + \displaystyle\int_a^b g(x) dx$

The integral of a sum is the sum of the integrals.
$\displaystyle\int_a^b [f(x)-g(x)] dx = \displaystyle\int_a^b f(x) dx - \displaystyle\int_a^b g(x) dx$

The integral of a difference is the difference of the integrals.
$\displaystyle\int_a^b cf(x) dx= c \displaystyle\int_a^b f(x) dx$

for constant $c$ . The integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function.
$\displaystyle\int_a^b f(x) dx = \displaystyle\int_a^c f(x) dx + \displaystyle\int_c^b f(x) dx$

Although this formula normally applies when $c$ is between $a$ and $b$ , the formula holds for all values of $a$ , $b$ , and $c$ , provided $f(x)$ is integrable on the largest interval.

Example: Using the Properties of the Definite Integral

Use the properties of the definite integral to express the definite integral of $f(x)=-3x^3+2x+2$ over the interval $[-2,1]$ as the sum of three definite integrals.

Show Solution

Using integral notation, we have $\displaystyle\int_{-2}^1 (-3x^3+2x+2) dx$ . We apply properties 3. and 5. to get

\begin{array}{ll} \int_{- 2}^{1} (- 3 x^{3} + 2 x + 2) d x & = \int_{- 2}^{1} - 3 x^{3} d x + \int_{- 2}^{1} 2 x d x + \int_{- 2}^{1} 2 d x \\ = - 3 \int_{- 2}^{1} x^{3} d x + 2 \int_{- 2}^{1} x d x + \int_{- 2}^{1} 2 d x \end{array}

$\begin{array}{ll} \displaystyle\int_{-2}^1 (-3x^3+2x+2) dx & = \displaystyle\int_{-2}^1 -3x^3 dx + \displaystyle\int_{-2}^1 2x dx + \displaystyle\int_{-2}^1 2 dx \\ & =-3 \displaystyle\int_{-2}^1 x^3 dx + 2 \displaystyle\int_{-2}^1 x dx + \displaystyle\int_{-2}^1 2 dx \end{array}$

Try It

Use the properties of the definite integral to express the definite integral of $f(x)=6x^3-4x^2+2x-3$ over the interval $[1,3]$ as the sum of four definite integrals.

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$6 \displaystyle\int_1^3 x^3 dx - 4 \displaystyle\int_1^3 x^2 dx + 2 \displaystyle\int_1^3 x dx - \displaystyle\int_1^3 3 dx$

Hint

Use the solving strategy from the previous example and the properties of definite integrals.

example: Using the Properties of the Definite Integral

If it is known that $\displaystyle\int_0^8 f(x) dx = 10$ and $\displaystyle\int_0^5 f(x) dx = 5$ , find the value of $\displaystyle\int_5^8 f(x) dx$ .

Show Solution

By property 6.,

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x

$\displaystyle\int_a^b f(x) dx = \displaystyle\int_a^c f(x) dx + \displaystyle\int_c^b f(x) dx$ .

Thus,

\begin{array}{lll} \int_{0}^{8} f (x) d x & = & \int_{0}^{5} f (x) d x + \int_{5}^{8} f (x) d x \\ 10 & = & 5 + \int_{5}^{8} f (x) d x \\ 5 & = & \int_{5}^{8} f (x) d x \end{array}

$\begin{array}{lll} \displaystyle\int_0^8 f(x) dx & = & \displaystyle\int_0^5 f(x) dx + \displaystyle\int_5^8 f(x) dx \\ 10 & = & 5 + \displaystyle\int_5^8 f(x) dx \\ 5 & = & \displaystyle\int_5^8 f(x) dx \end{array}$

Watch the following video to see the worked solution to Example: Using the Properties of the Definite Integral.

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Try It

If it is known that $\displaystyle\int_1^5 f(x) dx = -3$ and $\displaystyle\int_2^5 f(x) dx = 4$ , find the value of $\displaystyle\int_1^2 f(x) dx$ .

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Comparison Properties of Integrals

A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say that if a function $f(x)$ is above another function $g(x)$ , then the area between $f(x)$ and the $x$ -axis is greater than the area between $g(x)$ and the $x$ -axis. This is true depending on the interval over which the comparison is made. The properties of definite integrals are valid whether [latex]ab[/latex]. The following properties, however, concern only the case $a \le b$ , and are used when we want to compare the sizes of integrals.

Comparison Theorem

If $f(x) \ge 0$ for $a \le x \le b$ , then
$\displaystyle\int_a^b f(x) dx \ge 0$ .
If $f(x) \ge g(x)$ for $a \le x \le b$ , then
$\displaystyle\int_a^b f(x) dx \ge \displaystyle\int_a^b g(x) dx$ .
If $m$ and $M$ are constants such that $m \le f(x) \le M$ for $a \le x \le b$ , then
$m(b-a) \le \displaystyle\int_a^b f(x) dx \le M(b-a)$ .

example: Comparing Two Functions over a Given Interval

Compare $f(x)=\sqrt{1+x^2}$ and $g(x)=\sqrt{1+x}$ over the interval $[0,1]$ .

Show Solution

Graphing these functions is necessary to understand how they compare over the interval $[0,1]$ . Initially, when graphed on a graphing calculator, $f(x)$ appears to be above $g(x)$ everywhere. However, on the interval $[0,1]$ , the graphs appear to be on top of each other. We need to zoom in to see that, on the interval $[0,1], \, g(x)$ is above $f(x)$ . The two functions intersect at $x=0$ and $x=1$ (Figure 9).

A graph of the function f(x) = sqrt(1 + x^2) in red and g(x) = sqrt(1 + x) in blue over [-2, 3]. The function f(x) appears above g(x) except over the interval [0,1]. A second, zoomed-in graph shows this interval more clearly.

Figure 9. (a) The function $f(x)$ appears above the function $g(x)$ except over the interval $[0,1]$ (b) Viewing the same graph with a greater zoom shows this more clearly.

We can see from the graph that over the interval $[0,1], \, g(x) \ge f(x)$ . Comparing the integrals over the specified interval $[0,1]$ , we also see that $\displaystyle\int_0^1 g(x) dx \ge \displaystyle\int_0^1 f(x) dx$ (Figure 10). The thin, red-shaded area shows just how much difference there is between these two integrals over the interval $[0,1]$ .

A graph showing the functions f(x) = sqrt(1 + x^2) and g(x) = sqrt(1 + x) over [-3, 3]. The area under g(x) in quadrant one over [0,1] is shaded. The area under g(x) and f(x) is included in this shaded area. The second, zoomed-in graph shows more clearly that equality between the functions only holds at the endpoints.

Figure 10. (a) The graph shows that over the interval $[0,1], \, g(x) \ge f(x)$ , where equality holds only at the endpoints of the interval. (b) Viewing the same graph with a greater zoom shows this more clearly.

Watch the following video to see the worked solution to Example: Comparing Two Functions over a Given Interval.

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Module 5: Integration