Properties of Definite Integrals Calculus 1 AB

Properties of Definite Integrals Calculus 1 AB


Bam!!! Mr. Tarrou. In this video we are going
to be talking about properties of definite integrals. And starting off we are going to
look at the definition of two special definite integrals. That is is if function f is defined
at a value of x=a, then we define the definite integral of f(x) dx from a to a is equal to
zero. Now you can use definite integrals for other things, but if you are doing it for
area and you think of my lower and my upper bounds being pushed together and being the
same number, what is left in-between? Nothing So it is zero. If function f is integrable
on the closed interval from a to b, then we define the definite integral of f(x) dx from
b to a is going to be equal to… the same function f(x) dx… still doing a definite
integral… but see our lower and upper bounds have been swapped. We have b to a and now
we have a to b. Those variables or those values have switched places. We are maybe instead
of reading the axis left to right we are reading it from right to left. Well when you do that,
when you take those lower and upper bounds and swap places, you are going to get the
opposite answer. So here is an answer from a previous video I did. The definite integral
of -x^2+6 dx from the closed interval from 0 to 2 gave us a value of 28/3. We were finding
area at the time. The area was 28/3 square units. Well if I swapped those values and
make my lower bound instead of zero make it two, which doesn’t seem right because actually
2 is greater than 0 but if we swap those variables of a and b, then we are going to get what
kind of answer? Well… this says that if we swap a and b we are going to get opposite
answers. So instead of positive 28/3, we are going to get negative 28/3. The additive interval
property, if f is integrable on the 3 closed intervals determined by a, b, and then this
extra value of c that we are bringing into the problem…. and here is the picture of
what we are talking about. I have just drawn a function. I have our value of a and b marked
off on this case the x axis. I am bringing in a third variable. Kind of like you have
an 8 foot board and you cut it into a 2 foot and a 6 foot section. 2 and 6 still make 8
total feet. That is kind of what we are talking about here. If I take this area of a to b,
and I bring down another variable of c it now has 3 intervals. Now where are the 3 closed
intervals? There is one from a to c, there is one from c to b, and then it has to be
closed from a to b. That is where the 3 closed intervals are coming from even though you
might glance at this initially and think there is only two. So it says that the definite
integral of f(x) dx from a to b is equal to the definite integral f(x) dx from a to c
and then plus the definite integral from c to b of f(x) dx. So you know… do these two
parts in this case since we are looking… I have given us a diagram to look at in terms
of area… Does this area plus this area equal the entire area? It seems to make sense. And
we have for this diagram here, we would have the… actually I don’t need to write this.
It is right here. This situation exactly fits what we see here in our property. But it is
not absolutely necessary for c to lie in-between the values of a and b. Let me explain what
I mean by that. I have marked off this value of c here and it is still would be true that…
let’s see here. I am going to write this down. The definite integral from a to b of f(x)
dx… So this area here from a to b… is equal to the definite integral from a to c
of f(x) dx… that means that I am finding in this case, and again I am talking about
area, I have drawn this so that it will have a positive area from a to c. We want to find
the definite integral from a to c. Well a to c is way way out here. So then you go ok,
this property says that this definite integral should be equal to the integral here plus
the definite integral from c to b of f(x) dx. So it seems like I am already finding
a definite integral that is bigger than the green shaded area and then you want to add
even more area. If I take these value of b and c, and if you were to read this left to
right you would go a, b, and then you would get to c if you were to read those three variables
left to right. So I am going from a to c, and I am going back to b. Now let me use this
property here to rewrite this second term. The definite integral from a to c of f(x)
dx. If I swapped b and c, I bring the b down here, then by swapping those two variables
what I am going to get? The answer from this and the answer from this definite integral
are going to be opposites. So when I take the c and b and I switch places, it actually
is going to come out to be subtraction. Now that should make sense right. If I just want
the area of this green shade region and I find that definite integral from a to c, aren’t
I then going to need to subtract this area here because I have gone too far? Well that
is indeed what we are going to need to do. So as you look at the Additive Interval Property,
it is necessary for c to be in that enclosed interval of a to b. So let’s get to that next
screen. BAM! If f and g, functions f and g, are integrable on the closed interval of a
to b, and k is a constant then the functions k times f, and f plus or minus g are integrable
on that same closed interval of [a,b]. That math notation is the definite integral of
k*f(x)dx from a to b is equal to k times the definite integral of f(x) dx from a to b.
We can just take that constant that is in front of the function and move it out in front
of the definite integral notation. If we have a definite integral of two functions being
added or subtracted again with the dx and closed interval from [a,b], then we take the
definite integral of f, the definite integral of g, and then add that integration process….
either add or subtract our answers. So what does that mean really? Why do we care about
this? Because here we have the definite integral of -x^2+6x+8 dx on the closed interval of
1 to 4. Well this is a pretty complicated question. Certainly a lot more complicated
than the ones that I just did in the previous video Finding Areas of Common Geometric Regions.
However, this property here can allow us to take this quadratic, basically it is a parabola
opening down, and split it up into small manageable parts instead of going through the Reimann
Sum process or the Definition of Area… Or the limit of sums process. Whatever your book
is calling it. So what we are going to do is, we are going to rewrite this giving each
term its own definite integral. That means that we are going to have the definite integral
from 1 to 4 of -x^2 plus the definite integral from 1 to 4 of 6x plus the definite integral
of 1 to 4…. oops I forgot all of the dx’s… Getting all exited. plus the definite integral
from 1 to 4 of 6x dx plus the definite integral from 1 to 4 of 8 dx. Now do you see what I
just did there? I just came up with long problem and made up three little small integrals.
Now this is the definite integral of a constant, so we were to sketch this you would see that
we were just going to find the area of a rectangle. This is has a degree of one, so this is just
going to be a straight line with a slope of 6. My guess is… well this either going to
be in the form of when we sketch it we are either going to get a triangle or a trapezoid.
I think it is going to be a trapezoid. Then here we have the definite integral of -x^2dx.
Which we can also, it is a little bit easier to deal with these definite integrals when
we are working with positive values. So I am going to take this negative and move it
out front which is what the first property is saying I can do. Actually as far as that
is concerned I can take this 6 out as well. But this is pretty easily drawn so I am going
to leave that in there. We are going to take this constant of negative one and pull it
out front. We have negative, or the opposite, of the definite integral from 1 to 4 of x
squared dx plus the definite integral from 1 to 4 of 6x dx plus the same with the 8 dx.
Now this definite integral we can answer very quickly with a sketch. This definite integral
we can also answer with a sketch. The first one here, the x squares, if I don’t give you
the answer to the particular definite integral then you are going to have to go through the
limit process or the Reimann Sum process. So I am going to give you the answer to this
term just to save you some time. Because actually in my textbook where they are having you practice
these problems involving the properties of definite integrals, they give you all of these
answers. You just need to be able to break the problem up into separate individual terms
and then use the information given. I want to review finding the area of these common
Geometric regions since I have only done one video on that so far. And I will give you
the answer to this first one here when we come back. I am going to step out, draw these
diagrams that we need to answer the second and third terms. And also give you the answer
to this first one. We will finish with our total answer and be done with this video.
Ok. So my second term was the definite integral of 6x dx. We are looking at the function f(x)
is equal to 6x. I have drawn that line. It has a slope of 6 and a y intercept of 0. And
I have marked off the shaded area which we are concerned with. As you can see, it is
the shape of a trapezoid. The area of a trapezoid is 1/2 times the height times b sub 1 plus
b sub 2. So we have 1/2 times our height, and remember the bases are the parallel sides,
so our height is going form 1 to 4… or 4-1 is equal to 3… My first base here. Let’s
just mark that off as 6 units long. Our second base which 4 times 6 is 24. You understand
I am using 4 because that is the upper limit of my definite integral. So 4 times 6 is 24.
We have 1/2 of 3 times 6+24 is 30. One-half of 30 is 15, and 15 times 3…. 15 times 3…
Well 10 times 3 is 30, 5 times 3 is 15, so that comes out to be 45 square units since
we are finding area. That means the definite integral of 6x dx is equal to 45. Off to the
side I have said the definite integral of x^2 dx from that closed interval from 1 to
4 is equal to 21. But we have the negative out here that we pulled from the definite
integral. So actually while this is equal to 21, we have that negative out front, so
let’s not forget that. And then finally of course our constant f(x) is… forgot to write
that down… f(x)=8. y=8 is a horizontal line. The definite integral between 1 and 4. So
we are just finding the area of that rectangle which is just length times width. So the area
is 3 times 8 which is 24 square units since we are finding area. What do we have? 24-21
is 3, and 3 plus 45 is 48 square units. That is the end of my discussion about properties
of definite integrals. That means it is time for me to get out of here. I am Mr. Tarrou.
BAM! Go Do Your Homework:)

25 thoughts on “Properties of Definite Integrals Calculus 1 AB

  1. Thank you for noticing…did I have my shirt on that says so in this video?!..hahahaLOL
    I wear funny shirts in alot of my videos and I do have one that says "I know I'm awesome, now enough about me" but it's green so I guess I just got lucky and had an awesome day:))…happy to help..click on my channel anytime!

  2. Your teaching methodology is excellent and impressive………! Hope every lectures are just like ur lectures……………..!

  3. Tags you can add for the video…  adding integrals, additive property of integrals, evaluating integrals…  🙂  

  4. Hi 
    I think u forgot to mention the max-min inequality for definite integrals .
    I really don't get this property . 
    Thank u 

  5. Sir, why do we use graphs instead we could directly find the integrals by general rule, waiting….

  6. It is nicely done, i must say. Very readable and level of understanding is high. But, i have a question..
    You explained how you got 45 by drawing and calculating the trapezoid with formulas, but just by calculating the integral i got the value 47.5. How come is that, i wonder?

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