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Properties of the Definite Integral

Author: Sophia
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what's covered
In this lesson, you will learn some useful properties of definite integrals. They are mostly quite intuitive and have been hinted at in previous tutorials, but this is where we will establish these properties officially. Specifically, this lesson will cover:

Table of Contents

1. Properties of the Definite Integral

formula to know
Property Integral Formula In Words
1 Definite Integral When Lower and Upper Bounds Are Equal

integral subscript a superscript a f open parentheses x close parentheses d x equals 0 		When the limits of integration are equal, the value of the definite integral is 0.
2 Definite Integral When Upper and Lower Bounds Are Interchanged

integral subscript b superscript a f open parentheses x close parentheses d x equals short dash integral subscript a superscript b f open parentheses x close parentheses d x 		When the order of the limits of integration are interchanged, the values of the definite integrals are opposites.
3 Definite Integral of a Constant Function

integral subscript a superscript b k d x equals k open parentheses b minus a close parentheses 		The definite integral of a constant is equal to the constant multiplied by the width of the interval open parentheses b minus a close parentheses.
4 Definite Integral of a Constant Multiple of a Function

integral subscript a superscript b k times f open parentheses x close parentheses d x equals k integral subscript a superscript b f open parentheses x close parentheses d x 		The constant k can be moved outside, and the definite integral of f open parentheses x close parentheses is multiplied by k.
5 Definite Integral Over a Partition of an Interval, with a ≤ b ≤ c

integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript b superscript c f open parentheses x close parentheses d x equals integral subscript a superscript c f open parentheses x close parentheses d x 		Adding areas
Note: a less or equal than b less or equal than c

To understand some of these properties, we’ll look at Riemann sums and areas.

Property 1: If the lower and upper limits of integration are equal, then the width of the interval is 0, which means there is no accumulated area.

Property 2: As we have seen, integral subscript a superscript b f open parentheses x close parentheses d x accumulates area from x equals a to x equals b. Then, moving in the opposite direction from x equals b to x equals a comma whatever was added would be subtracted, and vice versa. Thus, the definite integrals have opposite signs.

Property 3: Consider the graph of y equals k on the interval open square brackets a comma space b close square brackets.

A graph with an x-axis with labels ‘a’ and ‘b’ at two points, where ‘b’ is greater than ‘a’, and a y-axis with label ‘k’ at one point above the origin. A horizontal line labeled y equals k starts from the point (0, k) on the y-axis. Two vertical lines rise from x equals a and x equals b to meet the line y equals k. The area below the line y equals k and between x equals a and x equals b is shaded.

Assume k greater than 0 comma as in the picture. The region formed by y equals k comma x equals a comma and x equals b is a rectangle with height k and width b minus a. Then, the area is k open parentheses b minus a close parentheses.

  • When k greater than 0 comma the definite integral is equal to the area.
  • When k less than 0 comma k open parentheses b minus a close parentheses less than 0 comma which is also true since then the rectangle is below the x-axis.
Property 4: Consider the Riemann sums of each function, where the same partition is used:

Riemann Sum for bold italic y bold equals bold italic f open parentheses bold x close parentheses Riemann Sum for bold italic y bold equals bold italic k bold times bold italic f open parentheses bold x close parentheses
sum from i equals 1 to n of f open parentheses c subscript i close parentheses increment x
Height of each rectangle: f open parentheses c subscript i close parentheses
Area of each rectangle: f open parentheses c subscript i close parentheses times increment x 		sum from i equals 1 to n of k times f open parentheses c subscript i close parentheses increment x
Height of each rectangle: k times f open parentheses c subscript i close parentheses
Area of each rectangle: k times f open parentheses c subscript i close parentheses times increment x

In the Riemann sum, all terms have a common factor of k, meaning it can be factored outside the sum, k times sum from i equals 1 to n of f open parentheses c subscript i close parentheses increment x. Since this is the original Riemann sum multiplied by k, this justified the integral version of this property.

Property 5: Consider the graph in the figure:

A graph with an x-axis represents the function y equals f(x). The x-axis has three labeled points ‘a’, ‘b’, and ‘c’, where b is greater than a and c is greater than b, defining two intervals. Three vertical lines rise from (a, 0), (b, 0), and (c, 0). The line x equals c is longer than the lines x equals a and x equals b. The vertical lines at x equals a and x equals b are approximately the same height. A curve starts from the top of the vertical line x equals a, rises to a peak, and then dips downward to the line at x equals b. From there, the curve continues to decline for a while before rising to another peak just before the vertical line at x equals c and then dips downward to the line x equals c. The area under the curve from x equals a to x equals c is shaded.

By adding areas, we see that the area on open square brackets a comma space c close square brackets is the sum of the areas on open square brackets a comma space b close square brackets and open square brackets b comma space c close square brackets.

EXAMPLE

The graph in the figure shows a function f open parentheses x close parentheses and areas between f open parentheses x close parentheses and the x-axis.

A graph with an x-axis represents the function y equals f(x). The x-axis ranges from 0 to 7. A curve steeply rises from point 0 on the x-axis, dips downward, then rises again slightly and dips downward, and crosses the x-axis at the point (3, 0). The curve then continues downward and rises to cross the x-axis again at the point (5, 0), steeply rises to a peak, dips downward, and crosses the x-axis at (6, 0). From here, the curve then continues downward and then rises to meet the x-axis at (7, 0). The shaded areas below the curve between (0, 0) and (3, 0), above the curve between (3, 0) and (5, 0), below the curve between (5, 0) and (6, 0), and above the curve between (6, 0) and (7, 0) are labeled ‘5’, ‘2’, ‘2’, and ‘1’, respectively, each indicated by an arrow.
Find the definite integrals of f for each of the following:

a. Find the definite integral: integral subscript 3 superscript 5 f open parentheses x close parentheses d x

integral subscript 3 superscript 5 f open parentheses x close parentheses d x Evaluate this definite integral.
equals short dash 2 The area of the region is 2, but is below the x-axis.

Conclusion: integral subscript 3 superscript 5 f open parentheses x close parentheses d x equals short dash 2

b. Find the definite integral: integral subscript 0 superscript 5 f open parentheses x close parentheses d x

integral subscript 0 superscript 5 f open parentheses x close parentheses d x Evaluate this definite integral.
equals integral subscript 0 superscript 3 f open parentheses x close parentheses d x plus integral subscript 3 superscript 5 f open parentheses x close parentheses d x Use the following property: integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript b superscript c f open parentheses x close parentheses d x equals integral subscript a superscript c f open parentheses x close parentheses d x
Note, “3” was chosen since at x equals 3 comma the first region ends and the second one begins.
equals 5 plus open parentheses short dash 2 close parentheses Substitute values from the graph:
integral subscript 0 superscript 3 f open parentheses x close parentheses d x equals 5 comma space integral subscript 3 superscript 5 f open parentheses x close parentheses d x equals short dash 2
equals 3 Simplify.

Conclusion: integral subscript 0 superscript 5 f open parentheses x close parentheses d x equals 3

c. Find the definite integral: integral subscript 5 superscript 3 f open parentheses x close parentheses d x

integral subscript 5 superscript 3 f open parentheses x close parentheses d x Evaluate this definite integral.
equals short dash integral subscript 3 superscript 5 f open parentheses x close parentheses d x Use the following property: integral subscript b superscript a f open parentheses x close parentheses d x equals short dash integral subscript a superscript b f open parentheses x close parentheses d x
Since the limits of integration are in reverse order, this property is appropriate to use.
equals short dash open parentheses short dash 2 close parentheses equals 2 Substitute values from the graph:
integral subscript 3 superscript 5 f open parentheses x close parentheses d x equals short dash 2
equals 2 Simplify.

Conclusion: integral subscript 5 superscript 3 f open parentheses x close parentheses d x equals 2

d. Find the definite integral: integral subscript 0 superscript 5 4 f open parentheses x close parentheses d x

integral subscript 0 superscript 5 4 f open parentheses x close parentheses d x Evaluate this definite integral.
equals 4 integral subscript 0 superscript 5 f open parentheses x close parentheses d x Use the following property: integral subscript a superscript b k times f open parentheses x close parentheses d x equals k integral subscript a superscript b f open parentheses x close parentheses d x
equals 4 open parentheses 3 close parentheses integral subscript 0 superscript 5 f open parentheses x close parentheses d x was evaluated in part b.
equals 12 Simplify.

Conclusion: integral subscript 0 superscript 5 4 f open parentheses x close parentheses d x equals 12

try it
Consider the following definite integrals in relation to the same graph as in the previous example.
a. integral subscript 4 superscript 4 open square brackets f open parentheses x close parentheses close square brackets cubed d x
b. integral subscript 5 superscript 7 short dash 3 f open parentheses x close parentheses d x
c. integral subscript 3 superscript 6 f open parentheses x close parentheses d x
Evaluate each definite integral.
a. Since the lower and upper limits are equal, the value of the definite integral is 0.

b. From properties of integrals, integral subscript 5 superscript 7 short dash 3 f open parentheses x close parentheses d x equals short dash 3 integral subscript 5 superscript 7 f open parentheses x close parentheses d x.

From the graph, integral subscript 5 superscript 7 f open parentheses x close parentheses d x equals 2 minus 1 equals 1 since the region on open square brackets 5 comma space 6 close square brackets has area 2 above the x-axis, and the region on open square brackets 6 comma space 7 close square brackets has area 1 below the x-axis.

Then, short dash 3 integral subscript 5 superscript 7 f open parentheses x close parentheses d x equals short dash 3 open parentheses 1 close parentheses equals short dash 3.

c. To find integral subscript 3 superscript 6 f open parentheses x close parentheses d x comma consider integral subscript 3 superscript 5 f open parentheses x close parentheses d x and integral subscript 5 superscript 6 f open parentheses x close parentheses d x.

From the graph:

  • integral subscript 3 superscript 5 f open parentheses x close parentheses d x equals short dash 2 comma since the area of the region is 2, but below the x-axis.
  • integral subscript 5 superscript 6 f open parentheses x close parentheses d x equals 2 comma since the area of the region is 2, and above the x-axis.
Then, integral subscript 3 superscript 6 f open parentheses x close parentheses d x equals integral subscript 3 superscript 5 f open parentheses x close parentheses d x plus integral subscript 5 superscript 6 f open parentheses x close parentheses d x equals short dash 2 plus 2 equals 0.

2. Properties of Definite Integrals of Combinations of Functions

formula to know
Formula In Words
Definite Integral of a Sum of Two Functions

integral subscript a superscript b open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript a superscript b g open parentheses x close parentheses d x 		The definite integral of a sum of two functions is the sum of the definite integrals of the functions.
Definite Integral of a Difference of Two Functions

integral subscript a superscript b open square brackets f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x minus integral subscript a superscript b g open parentheses x close parentheses d x 		The definite integral of a difference of two functions is the difference of the definite integrals of the functions.

These properties follow directly from Riemann sums (using properties of summations).

For the sum property:

integral subscript a superscript b open parentheses f open parentheses x close parentheses plus g open parentheses x close parentheses close parentheses d x
equals limit as n rightwards arrow infinity of open square brackets sum from k equals 1 to n of open parentheses f open parentheses c subscript k close parentheses plus g open parentheses c subscript k close parentheses close parentheses increment x close square brackets
equals limit as n rightwards arrow infinity of open square brackets sum from k equals 1 to n of open parentheses f open parentheses c subscript k close parentheses increment x plus g open parentheses c subscript k close parentheses increment x close parentheses close square brackets

By the property of summations, this is written as limit as n rightwards arrow infinity of open square brackets sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x plus sum from k equals 1 to n of g open parentheses c subscript k close parentheses increment x close square brackets comma which by the limit property is equal to limit as n rightwards arrow infinity of sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x plus limit as n rightwards arrow infinity of sum from k equals 1 to n of g open parentheses c subscript k close parentheses increment x comma which is equal to integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript a superscript b g open parentheses x close parentheses d x.

A very similar sequence of steps can be followed for the difference between f and g.

EXAMPLE

Given integral subscript 2 superscript 6 f open parentheses x close parentheses d x equals 10 and integral subscript 2 superscript 6 g open parentheses x close parentheses d x equals 4 comma find each of the following:

a. Find the definite integral: integral subscript 2 superscript 6 open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x

integral subscript 2 superscript 6 open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x Evaluate this definite integral.
equals integral subscript 2 superscript 6 f open parentheses x close parentheses d x plus integral subscript 2 superscript 6 g open parentheses x close parentheses d x Use the definite integral of a sum of two functions property.
equals 10 plus 4 Substitute values.
equals 14 Simplify.

Conclusion: integral subscript 2 superscript 6 open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x equals 14

b. Find the definite integral: integral subscript 2 superscript 6 open square brackets 3 plus f open parentheses x close parentheses close square brackets d x

integral subscript 2 superscript 6 open square brackets 3 plus f open parentheses x close parentheses close square brackets d x Evaluate this definite integral.
equals integral subscript 2 superscript 6 3 d x plus integral subscript 2 superscript 6 f open parentheses x close parentheses d x Use the property:
integral subscript a superscript b open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript a superscript b g open parentheses x close parentheses d x
equals 3 open parentheses 6 minus 2 close parentheses plus 10 For the first integral, integral subscript a superscript b k d x equals k open parentheses b minus a close parentheses.
For the second integral, the value is given: 10
equals 22 Simplify.

Conclusion: integral subscript 2 superscript 6 open square brackets 3 plus f open parentheses x close parentheses close square brackets d x equals 22

c. Find the definite integral: integral subscript 2 superscript 6 open square brackets 3 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x

integral subscript 2 superscript 6 open square brackets 3 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x Evaluate this definite integral.
equals integral subscript 2 superscript 6 3 f open parentheses x close parentheses d x minus integral subscript 2 superscript 6 g open parentheses x close parentheses d x Use the property:
integral subscript a superscript b open square brackets f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x minus integral subscript a superscript b g open parentheses x close parentheses d x
equals 3 integral subscript 2 superscript 6 f open parentheses x close parentheses d x minus integral subscript 2 superscript 6 g open parentheses x close parentheses d x Use the property: integral subscript a superscript b k times f open parentheses x close parentheses d x equals k integral subscript a superscript b f open parentheses x close parentheses d x
equals 3 open parentheses 10 close parentheses minus 4 Substitute given values of integrals.
equals 26 Simplify.

Conclusion: integral subscript 2 superscript 6 open square brackets 3 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals 26

d. Find the definite integral: integral subscript 6 superscript 2 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x

Since we are given integral subscript 2 superscript 6 f open parentheses x close parentheses d x and integral subscript 2 superscript 6 g open parentheses x close parentheses d x comma let’s find integral subscript 2 superscript 6 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x first:

integral subscript 2 superscript 6 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x The integral we are trying to find, which is related to integral subscript 6 superscript 2 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x.
equals integral subscript 2 superscript 6 2 f open parentheses x close parentheses d x minus integral subscript 2 superscript 6 g open parentheses x close parentheses d x Use the property:
integral subscript a superscript b open square brackets f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x minus integral subscript a superscript b g open parentheses x close parentheses d x
equals 2 integral subscript 2 superscript 6 f open parentheses x close parentheses d x minus integral subscript 2 superscript 6 g open parentheses x close parentheses d x Use the property: integral subscript a superscript b k times f open parentheses x close parentheses d x equals k integral subscript a superscript b f open parentheses x close parentheses d x
equals 2 open parentheses 10 close parentheses minus 4 Given integral subscript 2 superscript 6 f open parentheses x close parentheses d x equals 10 and integral subscript 2 superscript 6 g open parentheses x close parentheses d x equals 4.
equals 16 Simplify.

Now, since integral subscript 2 superscript 6 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals 16 comma it follows that integral subscript 6 superscript 2 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals short dash 16. This is due to the property that integral subscript b superscript a f open parentheses x close parentheses d x equals short dash integral subscript a superscript b f open parentheses x close parentheses d x.

Conclusion: integral subscript 6 superscript 2 open square brackets 2 f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals short dash 16

watch
In this video, given integral subscript 1 superscript 4 f open parentheses x close parentheses d x equals 12 comma we’ll find the value of integral subscript 1 superscript 4 open parentheses f open parentheses x close parentheses plus x close parentheses d x.

3. Comparison Properties

3a. Comparing Two Functions

If f open parentheses x close parentheses less or equal than g open parentheses x close parentheses on open square brackets a comma space b close square brackets comma then integral subscript a superscript b f open parentheses x close parentheses d x less or equal than integral subscript a superscript b g open parentheses x close parentheses d x.

To visualize this, consider these graphs. Clearly, the area between the graph of g open parentheses x close parentheses and the x-axis is greater than the area between the graph of f open parentheses x close parentheses and the x-axis.

The graphs of bold italic f open parentheses bold x close parentheses and bold italic g open parentheses bold x close parentheses together on the same axes. The region bounded by the graphs of bold italic g open parentheses bold x close parentheses and the x-axis.
bold Area bold equals bold integral subscript bold 0 superscript bold 3 bold italic g open parentheses bold x close parentheses bold italic d bold italic x 		The region bounded by the graphs of bold italic f open parentheses bold x close parentheses and the x-axis.
bold Area bold equals bold integral subscript bold 0 superscript bold 3 bold italic f open parentheses bold x close parentheses bold italic d bold italic x
A graph with an x-axis ranging from 0 to 4 and a y-axis ranging from 0 to 11. The graph has a parabolic curve g(x) and a line f(x). The parabolic curve g(x) starts at the point (0, 2) and rises upward to the point (3, 11). The line f(x) slants upward from the point (0, 0) up to the point (3, 3). A graph with an x-axis ranging from 0 to 4 and a y-axis ranging from 0 to 11. The graph has a parabolic curve and a dashed line. The parabolic curve starts at the point (0, 2) and rises upward to the point (3, 11). A dashed line slants upward from the point (0, 0) up to the point (3, 3). The area under the parabolic curve down to the x-axis is shaded from x equals 0 to x equals 3. A graph with an x-axis ranging from 0 to 4 and a y-axis ranging from 0 to 11. The graph has a parabolic curve and a line. The parabolic curve starts at (0, 2) and rises upward to the point (3, 11). The line slants upward from the point (0, 0) to (3, 3). The area under the line and above the horizontal axis between the origin and x equals 3 is shaded.

3b. Bounds on the Value of a Definite Integral

Let m = the minimum value of f open parentheses x close parentheses on open square brackets a comma space b close square brackets.
Let M = the maximum value of f open parentheses x close parentheses on open square brackets a comma space b close square brackets.

Given that m less or equal than f open parentheses x close parentheses less or equal than M on the interval open square brackets a comma space b close square brackets comma then m open parentheses b minus a close parentheses less or equal than integral subscript a superscript b f open parentheses x close parentheses d x less or equal than M open parentheses b minus a close parentheses.

Below is the graphical justification:

Three graphs each labeled ‘a’ and ‘b’ on the horizontal axis such that b is greater than a, representing the function. Two dashed lines rise from point ‘a’ and point ‘b’ in all three graphs. All three graphs have a curve that rises from the upper left of the dashed line at x equals a, reaches a peak, dips to reach the inverted peak, and then finally rises to a point on the lower right corner of the line x equals b. In the first graph, the area between the dashed lines x equals a and x equals b up to the inverted peak of the curve is shaded, forming a rectangle. In the second graph, the area between the dashed lines x equals a and x equals b below the curve and above the x-axis is shaded. In the third graph, the area between the dashed lines x equals a and x equals b is shaded. The shaded region is labeled ‘min of f’ to the left of the dashed line x equals a in the first graph, and the shaded region of the third graph is labeled ‘max of f’ to the right of the dashed line x equals b. The horizontal axis of the graphs is labeled as follows: min of ‘f’ multiplied by b − a in simple brackets with an arrow pointing toward the first graph is less than equal to the integration of f(x) dx ranging from a to b with an arrow pointing toward the second graph, which is less than equal to the max of ‘f’ multiplied by b − a in simple brackets pointing toward the third graph.

EXAMPLE

Use the graph to determine the upper and lower bounds of the value of integral subscript 1 superscript 5 f open parentheses x close parentheses d x.

A graph with an x-axis ranging from 0 to 6 and a y-axis ranging from 0 to 9. The graph consists of a wavy curve representing the function y equals f(x) with gentle rises and dips. The curve rises from the point (1, 6), reaches the peak at (2, 9), and then dips in a wavy manner to reach a minimum at the point (4, 2). The curve then rises in a wavy manner to reach the point (5, 4).

The minimum value of f open parentheses x close parentheses on open square brackets 1 comma space 5 close square brackets is 2; the maximum value is 9.

Thus, the minimum value for the definite integral is 2 open parentheses 5 minus 1 close parentheses equals 8 and the maximum value is 9 open parentheses 5 minus 1 close parentheses equals 36. That is, 8 less or equal than integral subscript 1 superscript 5 f open parentheses x close parentheses d x less or equal than 36.

try it
Consider the graph below in relation to the integral integral subscript short dash 2 end subscript superscript 3 f open parentheses x close parentheses d x.

A graph with an x-axis ranging from –2 to 3 and a y-axis ranging from 0 to 1. A curve rises from the marked point at (–2, 0.2) to the marked point at (0, 1) and then dips downward to the marked point at (3, 0.1).

Use the graph to determine the upper and lower bounds of the value of this integral.
Note that the length of the interval is 3 minus open parentheses short dash 2 close parentheses equals 5.

Since the lowest value of f open parentheses x close parentheses on the interval open square brackets short dash 2 comma space 3 close square brackets is 0.1, the smallest possible value of integral subscript short dash 2 end subscript superscript 3 f open parentheses x close parentheses d x equals 0.1 open parentheses 5 close parentheses equals 0.5.

Similarly, since the highest value of f open parentheses x close parentheses on the interval open square brackets short dash 2 comma space 3 close square brackets is 1, the largest possible value of integral subscript short dash 2 end subscript superscript 3 f open parentheses x close parentheses d x equals 1 open parentheses 5 close parentheses equals 5.

Therefore, the value of integral subscript short dash 2 end subscript superscript 3 f open parentheses x close parentheses d x is between 0.5 and 5.

summary
In this lesson, you examined some useful properties of the definite integral as well as properties of definite integrals of combinations of functions. As a result of properties of areas, you are now able to find definite integrals of sums, differences, and constant multiples of functions, as well as utilize comparison properties to compare two functions and put bounds on the value of a definite integral.

Source: THIS TUTORIAL HAS BEEN ADAPTED FROM CHAPTER 4 OF "CONTEMPORARY CALCULUS" BY DALE HOFFMAN. ACCESS FOR FREE AT WWW.CONTEMPORARYCALCULUS.COM. LICENSE: CREATIVE COMMONS ATTRIBUTION 3.0 UNITED STATES.

Formulas to Know
Definite Integral Over a Partition of an Interval, with a ≤ b ≤ c

integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript b superscript c f open parentheses x close parentheses d x equals integral subscript a superscript c f open parentheses x close parentheses d x

Definite Integral When Lower and Upper Bounds Are Equal

integral subscript a superscript a f open parentheses x close parentheses d x equals 0

Definite Integral When Upper and Lower Bounds Are Interchanged

integral subscript b superscript a f open parentheses x close parentheses d x equals short dash integral subscript a superscript b f open parentheses x close parentheses d x

Definite Integral of a Constant Function

integral subscript a superscript b k d x equals k open parentheses b minus a close parentheses

Definite Integral of a Constant Multiple of a Function

integral subscript a superscript b k times f open parentheses x close parentheses d x equals k integral subscript a superscript b f open parentheses x close parentheses d x

Definite Integral of a Difference of Two Functions

integral subscript a superscript b open square brackets f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x minus integral subscript a superscript b g open parentheses x close parentheses d x

Definite Integral of a Sum of Two Functions

integral subscript a superscript b open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets d x equals integral subscript a superscript b f open parentheses x close parentheses d x plus integral subscript a superscript b g open parentheses x close parentheses d x

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