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.

## 14 Comments

## Atchu K

Good sir

## Muneef Qureshi

I want to ask a question that in reaction of sodium and chlorine to make salt how can chlorine react with sodium when it already cl2

## Krish Mehta

Some men just want the world to learn…

## Thaumius

Dave, When you intergrate something don't you have to put a constant C? or is it only if the range is not specified?

## Minespidur

Thanks so much!

## Ayush Shukla

How can I contract you

## Bruno Volpato

Thanks!

## Hamid Alrawi

thank you professor Dave, your math videos are very good and useful as much as your chemistry videos <3

## Joshua Michael Saberon

so much appreciated! Thank you Professor Dave!

## Deepankar Chakraborty

Amazing Explanation… Also, Thank You for responding to my Email.

## Owam Payiya

First of all youre great. Thats it

## Yat Sum Leung

king

## Aditya Shankar

Hey prof! What'll be Antiderivative of x^-1

## El Prufesur

the best