How to Understand Calculus: Integration Rules and Examples
What's Covered in this Tutorial
In this second part of a two part tutorial, we cover:
- Concept of integration
- Definition of indefinite and definite integrals
- Integrals of common functions
- Rules of integrals and worked examples
- Applications of integral calculus, volumes of solids, real world examples
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Integration is a Summing Process
We saw in the first part of this tutorial how differentiation is a way of working out the rate of change of functions. Integration in a sense is the opposite of that process. It is a summing process used to add up infinitesimally small quantities.
What is Integral Calculus Used For?
Integration is a summing process, and as a mathematical tool it can be used for:
- evaluating the area under functions of one variable
- working out the area and volume under functions of two variables or summing multidimensional functions
- calculating the surface area and volume of 3D solids
In science, engineering, economics etc, real world quantities such as temperature, pressure, magnetic field strength, illumination, speed, flow rate, share values etc can be described by mathematical functions.Integration allows us to integrate these variables to arrive at a cumulative result.
Area Under a Graph of a Constant Function
Imagine we have a graph showing the velocity of a car versus time. The car travels at a constant velocity of 50 mph, so the plot is just a horizontal straight line.
The equation for distance travelled is:
Distance travelled = velocity x time
So in order to calculate distance travelled at any point in the journey, we multiply the height of the graph (the velocity) by the width (time) and this is just the rectangular area under the graph of velocity. We are integrating velocity to calculate distance. The resulting graph we produce for distance versus time is a straight line.
So if the car's velocity is 50 mph, then it travels
50 miles after 1 hour
100 miles after 2 hours
150 miles after 3 hours
200 miles after 4 hours and so on.
Note that an interval of 1 hour is arbitrary, we can choose it to be anything we want.
If we draw a graph of distance travelled versus time, we see how the distance increases with time. The graph is a straight line.
Area Under a Graph of a Linear Function
Now let's make things a little bit more complicated!
This time we'll use the example of filling a water tank from a pipe.
Initially there is no water in the tank and no flow into it, but over a period of minutes, the flow rate increases continuously.
The increase in flow is linear which means that the relationship between flow rate in gallons per minute and time is a straight line.
We use a stopwatch to check elapsed time and record the flow rate every minute. (Again this is arbitrary).
After 1 minute, flow has increased to 5 gallons per minute.
After 2 minutes, flow has increased to 10 gallons per minute.
and so on.....
Flow rate is in gallons per minute (gpm) and volume in the tank is in gallons.
The equation for volume is simply:
Volume = average flow rate x time
Unlike the example of the car, to work out the volume in the tank after 3 minutes, we can't just multiply the flow rate (15 gpm) by 3 minutes because the rate was not at this rate for the full 3 minutes. Instead we multiply by the average flow rate which is 15/2 = 7.5 gpm.
So volume = average flow rate x time = (15/2) x 3 = 2.5 gallons
In the graph below, this just turns out to be the area of the triangle ABC.
Just like the car example, we are calculating the area under the graph.
If we record the flow rate at intervals of 1 minute and work out the volume, the increase in water volume in the tank is an exponential curve.
What is Integration?
It is a summing process used to add up infinitesimally small quantities
Now consider a case where the flow rate into the tank is variable and non-linear. Again we measure the flow rate at regular intervals. Just like before, the volume of water is the area under the curve. We can't use a single rectangle or triangle to calculate area, but we can try to estimate it by dividing it up into rectangles of width Δt, calculating the area of those and summing the result. However there will be errors and the area will be underestimated or over estimated depending on whether the graph is increasing or decreasing.
Using Numerical Integration to Find the Area Under a Curve.
We can improve accuracy by making the intervals Δt shorter and shorter.
We are in effect using a form of numerical integration to estimate the area under the curve by adding together the area of a series of rectangles.
Now consider a general function y = f (x).
We are going to specify an expression for the total area under the curve over a domain [a,b] by summing a series of rectangles. In the limit, the width of the rectangles will become infinitesimally small and approach 0. The errors will also become 0.
- The result is called the definite integral of f (x) over the domain [a,b].
- The ∫ symbol means "the integral of" and the function f (x) is being integrated.
- f (x) is called an integrand.
The sum is called a Riemann Sum. The one we use below is called a right Reimann sum. dx is an infinitesimally small width. Roughly speaking, this can be thought of as the value Δx becomes as it approaches 0. The Σ symbol means that all the products f (xi)xi (the area of each rectangle) are being summed from i = 1 to i = n and as Δx → 0, n → ∞.
The Difference Between Definite and Indefinite Integrals
Analytically we can find the anti-derivative or indefinite integral of a function f (x).
This function has no limits.
If we specify an upper and lower limit, the integral is called a definite integral.
Using Indefinite Integrals to Evaluate Definite Integrals
If we have a set of data points, we can use numerical integration as described above to work out the area under curves. Although it wasn't called integration, this process has been used for thousands of years to calculate area and computers have made it easier to do the arithmetic when thousands of data points are involved.
However if we know the function f (x) in equation form (e.g. f (x) = 5x2 + 6x +2) , then firstly knowing the anti-derivative (also called the indefinite integral) of common functions and also using rules of integration, we can analytically work out an expression for the indefinite integral.
The fundamental theorem of calculus then tells us that we can work out the definite integral of a function f(x) over an interval using one of its anti-derivatives F(x). Later we will discover that there are an infinite number of anti-derivatives of a function f (x).
Indefinite Integrals and Constants of Integration
The table below shows some common functions and their indefinite integrals or anti-derivatives. C is a constant. There are an infinite number of indefinite integrals for each function because C can have any value.
Why is this?
Consider the function f (x) = x3
We know the derivative of this is 3x2
What about x3 + 5 ?
d/dx ( x3 + 5 ) = d/dx (x3) + d/dx(5) = 3x2 + 0 = 3x2 ....... the derivative of a constant is 0
So the derivative of x3 is the same as the derivative of x3 + 5 and = 3x2
What's the derivative of x3 + 3.2 ?
Again d/dx ( x3 + 3.2 ) = d/dx (x3) + d/dx(3.2) = 3x2 + 0 = 3x2
No matter what constant is added to x3, the derivative is the same.
Graphically we can see that if functions have a constant added, they are vertical translations of each other, so since the derivative is the slope of a function, this works out the same no matter what constant is added.
Since integration is the opposite of differentiation, when we integrate a function, we must add on a constant of integration to the indefinite integral
So e.g. d/dx( x3) = 3x2
and ∫ 3x2 dx = x3 + C
Indefinite Integrals of Common Functions
∫ a dx
ax + C
∫ x dx
x² / 2 + C
∫ 1/x dx
ln x + C
∫ x² dx
x³ / 3 + C
∫ sin (x) dx
- cos (x) + C
∫ cos (x) dx
sin (x) + C
∫ sec ² (x) dx
tan (x) + C
∫ e ^ x dx
e ^ x + C
∫ a ^ x dx
(a ^ x) / ln (a) + C
∫ ln (x) dx
xln (x) - x + C
In the table below, u and v are functions of x.
u ' is the derivative of u wrt x.
v ' is the derivative of v wrt x.
Rules of Integration
Multiplication by a constant rule
∫ au dx
a ∫ u dx
∫ ( u + v ) dx
∫ u dx + ∫ v dx
∫ ( u - v ) dx
∫ u dx - ∫ v dx
Power rule (n ≠ -1)
∫ (x ^ n) dx
x ^ (n + 1) / (n + 1) + C
Reverse chain rule or integration by substitution
∫ f(u) u ' dx
∫ f(u) du + C ................... Replace u'(x)dx by du and integrate wrt u, then substitute back for the value of u in terms of x in the evaluated integral.
Integration by parts
∫ u v dx
u ∫ v dx + ∫ u ' ( ∫ v dx) dx
Examples of Working Out Integrals
Evaluate ∫ 7 dx
∫ 7 dx =
7 ∫ dx ..........multiplication by a constant rule
= 7x + C
What is ∫ 5x4 dx
∫ 5x4 dx = 5 ∫x4 dx ....... using multiplication by a constant rule
= 5(x5/5) + C .......... using power rule
= x5 + C
Evaluate ∫ (2x3 + cos(x) ) dx
∫ (2x3 + 6cos(x) ) dx = ∫ 2x3 dx + ∫ 6cos(x) dx .....using the sum rule
= 2 ∫ x3 dx + 6 ∫ cos(x) dx ..........using the multiplication by a constant rule
= 2(x4/4) + C1 + 6(sin(x) + C2 .....using the power rule. C1 and C2 are constants.
C1 and C2 can be replaced by a single constant C, so:
∫ (2x3 + cos(x) ) dx = x4/2 + 6sin(x) + C
Work out ∫ sin2(x)cos(x) dx
- We can do this using the reverse chain rule ∫ f(u) u'(x) dx = ∫ f(u) du where u is a function of x
- We use this when we have an integral of a product of a function of a function and its derivative
sin2(x) = (sin x) 2
Our function of x is sin x so replace sin(x) by u giving us sin2(x) = f(u) = u2 and cos(x)dx by du
So ∫ sin2(x)cos(x) dx = ∫ u2du = u3/3 + C
Substitute u = sin(x) back into the result:
u3/3 + C = sin3(x)/3 + c
So ∫ sin2(x)cos(x) dx = sin3(x)/3 + c
Evaluate ∫ xex^2 dx
It looks as though we could use the reverse chain rule for this example because 2x is the derivative of the exponent of e which is x2. However we need to adjust the form of the integral first. So write ∫ xex^2 dx as 1/2 x ∫ 2xex^2 dx = 1/2 ∫ ex^2 (2x) dx
No we have the integral in the form ∫ f(u) u' dx where u = x2
So 1/2 ∫ ex^2 (2x) = 1/2 ∫ eu u' dx = 1/2 ∫ eu du
but the integral of the exponential function eu is itself, do
1/2 ∫ eu du = 1/2 eu
Substitute for u giving
1/2 eu = 1/2 ex^2
Evaluate ∫ 6/(5x + 3) dx
- For this, we can use the reverse chain rule again.
- We know that 5 is the derivative of 5x + 3.
Rewrite the integral so that 5 is within the integral symbol and in a format that we can use the reverse chain rule:
∫ 6/(5x + 3) dx = ∫ (6/5) 5/(5x + 3) dx = 6/5∫ 1/(5x + 3) 5dx
Replace 5x + 3 by u and 5dx by du
6/5∫ 1/(5x + 3)5dx = 6/5∫ (1/u) du
But ∫ (1/u) du = ln(u) + C
So substituting back 5x + 3 for u gives:
∫ 6/(5x + 3) dx = 6/5∫ (1/u) du = 6/5ln(5x + 3) + C = 1.2ln(5x + 3) + C
© 2019 Eugene Brennan