# Introduction to Integration – Properties of Definite Integrals

The main properties of integrals are shown below. We will go over each property in detail by doing some examples. Order of Integration

There are two parts to this property. We will cover each part separately.

Part 1

For a function that is above the x-axis, when we integrate from left to right, the area under the curve is positive.

However, let’s say that we integrate from right to left. In this case, the area under the curve is negative.

Essentially, when we switch the order of integration, the sign of the definite integral changes. Therefore, Solution: Let’s draw the curve and shade the area under the curve: We see that the area under the curve is a triangle:

A = – ½ * b * h = ½ * 5 * 5 = – 25/2 square units

Remember! We are integrating from 5 to 0 (right to left). As a result, this area will be negative.

Constant Multiple

If we are taking the integral of a constant c, the area under the curve would simply be the value of the constant multiplied by the length of the interval: Let’s prove this graphically: As we can see, the area under the curve is a square. The area of the square is 2 * 2 = 4 square units.

However, let’s say that we take the integral of a constant multiplied by a function. Simply take the integral of the function and multiply the result by the constant:  Find the area of the trapezoid:

A = 1/2 * h * (b1+b2)

A = 1/2 * 3 * (2+5)

A = 10.5 square units

Now let’s multiply this area by the constant:

2 * 10.5 = 21  Solution: Our constant is -3, so we will factor the -3 from the integrand: Similarly, we can find the integral of the difference of two functions by simply taking the integral of each function separately, then adding the results:  Find the area of the triangle:

A1 = 1/2* b*h = 1/2*3*9= 27/2

Let’s find the area of the second interval: Using the constant multiple property,

A2= 2* (3-0) = 2 * 3 = 6

Of course, graphing the curve and shading the region gives us a rectangle: Therefore, the area under the integral is If we divide the integral of a function into subintervals, the sum of the area of all the subintervals is equal to the integral of the function over the entire interval: Examine the diagram below: The area under the curve is split into two areas. The area of the entire region is equal to the sum of the two areas. The area under the curve for the entire interval is 9 square units.

Integrals of Symmetric Functions

Before we discuss integrating symmetric functions, we need to talk about the different types of symmetry.

Even functions are functions that are symmetrical about the y axis. That means that if the point (x,y) exists on the graph, then the point (-x,y) must also exist on the graph. Odd functions are functions that are symmetrical about the origin. That means that if a point (x,y) exists on the graph, then the point (-x,-y) must also exist on the function. If a function is even, meaning that it is symmetrical about the y-axis, then Solution: Let’s graph the function and shade the region: f(x) = |x| is an even function because it is symmetrical about the y axis.

To take the integral, you can take the integral from [0,3] and multiply by 2 (because the area under the curve of [-3,0] is identical). The area under the curve of [0,3] is a triangle:

A = 1/2*3*3= 9/2

Multiply the area by 2.

9/2 * 2 = 9 Solution: Let’s graph the function and shade the region: This is an odd function because the function is symmetrical about the origin.

Notice that the area of the two regions are the same. However, one is below the x axis (which makes it negative), and the other one is above the x axis (which makes it positive). Therefore, the sum of the two areas is zero. Since the net area is zero, Comparison Property

There are 2 parts of the comparison property. We will go over each part separately.

Part 1

If two functions, g(x) and f(x) exists on the coordinate plane on the interval [a,b], and g(x) > f(x), meaning that g(x) is above f(x), then Consider the diagram below: Notice that on the interval of [a,b] g(x) is positioned above f(x). Therefore, when taking the area under the curve of g(x) and f(x), the area under g(x) will be greater than the area under f(x).

Part 2

Let’s say we wanted to find the integral of a function: We can determine the value of A, but let’s say we were told to estimate the value of A. We can estimate the exact value of A by finding the lowest and highest possible value of A.

To determine the lower bound for the value of an integral, we have to take the minimum value of the function (called “m”) and multiply it by the length of the interval (b – a).

Lower limit of integral value = m * (b – a) = Amin

To determine the upper bound for the value of an integral, we have to take the maximum value of the function (called “M”) and multiply it by the length of the interval (b – a).

Upper limit of integral value = M * (b-a) = Amax

Therefore, If we take the integral of a function on the interval [a,b], the value must fall in between the lower and upper bounds: The diagram below demonstrates this property. Example 10: Determine the lower and upper bounds of the value of . Next, find the exact value of the integral. Solution:

Lower bound of the integral

Let’s find the lowest value of f(x), which is found at x = 4:

f(4) = 10

Let’s multiply this value by the length of the interval:

f(4) * (b – a) = 10 * (6 – 1) = 10* 5 = 50

Upper bound of the integral

Next, let’s find the highest value of f(x), which is found at both x = 2 and x = 6:

f(2) = f(6) = 40

Let’s multiply this value by the length of the interval:

40 * (b – a) = 40 * (6 – 1) = 40* 5 = 200

The lower bound of the integral is 50, and the upper bound is 200. Therefore, the exact value of the integral must fall between 50 and 200. Let’s determine this value.

Exact value of integral

The area under the integral can be divided into three trapezoids: Find the area of each trapezoid:

A1=1/2*1*(40+30) =35

A2=1/2*2*(10+40) =25

A3=1/2*2*(10+40) =25

The value of References