Professor Dave here, let’s take some integrals.
Hopefully we now understand the relationship between differentiation and integration.
These are inverse operations, in the sense that integration requires taking the antiderivative
of a function.
For basic functions, this is easy to do, but it gets extremely complicated as the functions
get more complex, and we will have to learn a number of different strategies to tackle
the tougher ones.
But that will come later.
For now, let’s just go over a few properties of integrals, and then try some simple examples.
As we said, it will be a good idea to quickly highlight some important properties of definite integrals.
First, if we have some integral of f of x dx, over the interval from a to b, then if
we switch the limits of integration, integrating instead from b to a, this will be the same
as the first integral but negative.
This is because we are integrating in the opposite direction.
Next, if both of the limits of integration are the same number, the integral will be
equal to zero, no matter what the function is.
This is because we are essentially asking about the area under a single point, which
would be like finding the area of one infinitely thin rectangle, which is just a line.
A line has no area, so this integral equals zero.
Similarly, if we have two integrals of the same function over adjacent intervals, one
from a to b and the other from b to c, the sum of these integrals is equal to the integral
of the function over the whole interval, from a to c.
This is the same as saying that the area of this section plus the area of this section
is equal to the total area under the curve.
So that covers some properties regarding the limits of integration.
Next, when integrating a constant, we will simply get that constant times the quantity
of the upper limit minus the lower limit.
So the integral of three from one to five, will be three times the quantity five minus
one, or twelve.
This makes sense if you think of the function y equals three.
If we chop this off at one and five, this is now just a rectangle with a constant height
of three, so multiplying by the difference of these limits is the same as calculating
base times height, which gives us the area of the rectangle.
Similarly, if we are taking the integral of a constant times some function, we can just
pull the constant out of the integral, and we get the constant times the integral of
the function.
Then, we have the property that says that the integral of a sum of functions over some
interval is equal to the sum of their integrals over the same interval.
So the integral of f of x plus g of x dx equals the integral of f of x dx plus the integral
of g of x dx.
The same goes for the difference of functions, we just change these plus signs to minus signs.
Let’s also quickly note that if the interval of a function that is being integrated is
above the x-axis, its integral will be positive.
If instead it is below the x-axis, it will be negative, as these are negative y values.
If it crosses the x-axis, with some portion above and some portion below, the integral
will be the net area, or the area above the axis minus the area below it.
Keeping these properties in mind, let’s move forward and practice evaluating definite integrals.
To start out, we will just integrate simple polynomials.
This involves taking the antiderivative of a function, as we learned in the previous tutorial.
Let’s try something like x squared plus one, dx, from zero to two.
First things first, let’s get the antiderivative of this function.
X squared will become x cubed over three, and one will become x over one, or simply x.
Then we evaluate the antiderivative at the limits of integration, so let’s write F
of two minus F of zero.
F of two will be one third times two cubed, plus two.
Two cubed is eight, so we have eight thirds, and let’s turn two into six thirds, so that
we can combine to get fourteen thirds.
Then for zero, everything just becomes zero, and we end up simply subtracting zero, so
the answer is fourteen thirds.
Not too bad, right?
Let’s do a couple more.
How about two x squared plus root x, from one to four.
In finding the antiderivative, x squared becomes x cubed over three, so let’s just put the
three out here to get two thirds x cubed.
Now for root x, it may seem like we can’t evaluate this, but remember, root x is just
x to the one half power.
So we can absolutely do exactly the same thing, we just end up with fractions.
We add one to the exponent to get x to the three halves, and then we divide by three halves.
Dividing by three halves is the same as multiplying by two thirds, so we get two thirds x to the
three halves.
Now we are ready to plug in some numbers.
Four cubed is sixty four, times two thirds is one hundred twenty eight thirds.
Four to the three halves power can be evaluated by first taking the square root to get two,
and then cubing to get eight.
Times two thirds is sixteen thirds, so we add these together to get one hundred forty
four thirds.
Evaluating for one will be a little easier, since we just end up with two thirds plus
two thirds.
That’s four thirds to be subtracted from one hundred forty four thirds, leaving us
with one hundred and forty thirds.
How about just one more.
Let’s try the quantity (x to the fifth plus two x cubed plus one) all over x squared,
evaluated from one to two.
This seems much trickier, because it’s a fraction, but we can very easily express this
as a regular polynomial, we just have to divide each term in the numerator by x squared.
For x to the fifth, that gives us x cubed.
For two x cubed, that gives us two x.
And for one, that gives us one over x squared, but let’s remember our rules of exponents
and instead express this as x to the negative two.
That will allow us to more easily find the antiderivative.
Now we are ready to do what we normally do.
X cubed becomes x to the fourth over four, two x becomes two x squared over two, or simply
x squared, and then for x to the negative two, remember we are still just adding one
to the exponent to get x to the negative one, and then we divide by that exponent, so we
put this over negative one.
So that means we subtract x to the negative one, which is really one over x.
So that’s our antiderivative.
Now we are ready to plug in some numbers.
So this will be our final answer.
Hopefully after computing some simple integrals, we can see the immense power of this method.
Prior to the fundamental theorem of calculus, finding areas and distances associated with
curvature was incredibly complex, and only brilliant mathematicians could figure out
how to do it.
Now with this simple algorithm of finding the antiderivative, anyone can do it.
Now don’t get too cocky, integration will get much harder than this, but before moving
forward, let’s check comprehension.