13.04 Properties of definite integrals

Let's now look at how to use integral calculus to calculate the areas under a curve.

From our previous lesson we know that if we are given a continuous function $f$f on the interval $\left[a,b\right]$[a,b], where $f\left(x\right)\ge0$f(x)≥0 for all $x$x in the interval, then the value of the area under the graph of $y=f\left(x\right)$y=f(x) from $x=a$x=a to $x=b$x=b is called the definite integral and is written:

That is, for a function such as $f\left(x\right)$f(x) shown in the diagram above, where $f\left(x\right)\ge0$f(x)≥0 on the interval $a\le x\le b$a≤x≤b, we have $\int_a^bf(x)dx=A$∫ba​f(x)dx=A.

What does the notation $\int_a^bf(x)dx$∫ba​f(x)dx entail mathematically?

Once we have calculated the anti-derivative, we use square brackets around this primitive function and place the upper and lower limits on the right bracket. This notation shows in one step both the integral and what we're getting ready to substitute into our anti-derivative.

Let's look at some examples of how this is done below.

 

Worked examples

Example 1

Evaluate

Example 2

Evaluate

Practice questions

Question 1

Question 2

Question 3

 

Definite integrals, signed area, and area

Up to this point, we have only considered the calculation of definite integrals in regions where all function values are positive and this represented the area under the under the curve over the given interval. What do definite integrals represent when the function is partly or fully below the horizontal axis?

Consider the function $f(x)=x^2+2x-3$f(x)=x2+2x−3 between $x=-3$x=−3and $x=1$x=1:

We can see this area is completely below the horizontal axis. Setting up the integral we get:

 

We can see that the value of the definite integral is negative. As the result of the definite integral has a sign depending on whether the area is above or below the $x$x-axis, we call this the signed area.

In this case, the actual area bounded by the curve and the $x$x-axis is equivalent to the absolute value of the integral. That is, the area is $\frac{32}{3}$323​ units².

 

The example below highlights the difference between the calculation of signed area and area.

 

Worked example

Example 3

For the function $y=(x+2)^3-1$y=(x+2)3−1:

(a) Evaluate the integral of the function between $x=-3$x=−3 and $x=0$x=0.

Think: We are asked to evaluate a definite integral over a given integral, hence, we are calculating the signed area.

Do: We evaluate the following integral between $x=-3$x=−3 and $x=0$x=0:

(b) Calculate the area bounded by the function and the $x$x axis between $x=-3$x=−3 and $x=0$x=0

Think: we want to calculate the area over a given integral. Hence, we need to take care with positive and negative regions.

Do: Sketching the function over the given interval will tell us if there are going to be in any negative regions.

Using the calculator or the given form of the equation - the graph could be obtained from $y=x^3$y=x3 by a horizontal translation to the left by $2$2 units and then a vertical translation of $1$1 unit downwards, we obtain the following graph:

The area between $x=-3$x=−3 and $x=-1$x=−1 is below the horizontal axis and between $x=-1$x=−1 and $x=0$x=0 is above. So we will split our integral into two sections and take the negative of the integral for the part under the axis, as follows:

Hence the area is $4\frac{3}{4}$434​ units².

Reflect: Notice that the value for the area calculated in part (b) is greater than the signed area. This is because the signed area is the sum of positive and negative regions, whereas calculating the area takes the absolute value of each region and sums them together. Note that the integral for the negative region was given a negative sign to make it positive. Taking the absolute value would have achieved the same result.

 

Practice questions

Question 4

Question 5

 

Properties of definite integrals

Definite integrals have a number of properties that we can make use of in order to simplify and solve problems involving area and integration. These properties are summarised below.

The above properties are useful in manipulating integrals to simplify a problem or solve a problem given partial information. Let's look briefly at each property:

1. End-points equal:

This property is can be shown from the definition of a definite integral:

This also concurs with the idea that the area under a point is zero.

2. Function multiplied by a constant:

Just as for differentiation, when a function is multiplied by a constant $k$k and integrated, it is is the same as multiplying the integral of the function by the constant.

3. Switch end-points:

If we reverse the interval of a definite integral, that is instead of integrating from $a$a to $b$b we integrate from $b$b to $a$a we will get the negative of the original integral. This can be shown to be true from the definition of a definite integral as follows:

4. Sum of functions:

If a definite integral is the sum or difference of two functions, then this is the same as adding or subtracting the individual integral of each function over the interval. As with differentiation this means we can integrate each part of a function separately.

5. Split interval:

This property is very useful to combine to integrals with a shared start/end point or to split a given integral into separate regions.

We can split an integral over a closed interval $\left[a,c\right]$[a,c] at the point $b$b where $b$b is within the closed interval, the sum of the integral of the parts equals the original integral.

This can be shown from the definition of a definite integral as follows:

For areas under a curve we can also visualise this property as:

We can see in this case, where $a\le b\le c$a≤b≤c and $f\left(x\right)\ge0$f(x)≥0, we have:

 

Practice questions

Question 6

Question 7

Question 8

 

Using the calculator

We can use our calculator to evaluate definite integrals and to find areas under curves. When finding the area bound by a function and the $x$x-axis over an interval where the graph has sections that fall below the $x$x-axis, we can choose to sketch the function and calculate the area of positive and negative sections separately before summing the absolute value of each section, or we can integrate the absolute value of the function. That is, the area between the function $f\left(x\right)$f(x) and the $x$x-axis on the interval $a\le x\le b$a≤x≤b is:

Select the brand of calculator you use below to work through an example of using a calculator for calculations concerning areas under curves and evaluating definite integrals.

Casio ClassPad

TI Nspire