Antiderivatives - Integration, Business Mathematics & Statistics Video Lecture - B Com

BAM!
Mr. Tarrou here.
Are you tired of doing derivatives?
Well I have got the cure for you.
It is called the antiderivative.
We are going to undo all those derivatives.
You know, do the inverse math operation.
We are going to do that through a concept
called integration which we are going to build
on.
This is just the start of building the concept
to allow you to do integration which is going
to undo that derivative process.
It will allow you to, with indefinite integration,
find a version of... or if a little extra
information is given to you, find the actual
original function that gave you this derivative.
Now we are going to need at a little bit of
information, which we are not going to deal
with in this video, to allow us to find the
actual original function that gave us this
derivative that we are trying to undo.
Without that extra information we are just
going to have the ability to find the General
Antiderivative which is what this video is
about.
Part 2 is going to be where we can find the
actual original function.
To explain what I am talking about here, I
am using some notation which I will change
in a second to match the definition that we
are going to deal with.
But we have... kind of like in the three chapters
of my book anyway that we do derivatives in
the beginning... we will just label this f(x)
and f'(x) indicating that we found the derivative.
So I have three functions here.
y=x^4, y=x^4+2, and y=x^4-5.
So three different original functions.
Now remembering that the first derivative
allows us to find the slope of a curve along
pretty much any given value of x where the
function is defined, well look at the plus
two and the minus five is.
When you have an addition of a constant at
the end of a function, it is only a vertical
shift.
So if I take a graph that maybe has a slope
here of 2, so the first derivative when you
plug in this particular value of x would be
equal to 2, if I take that function and I
just move it up a few places... or down...
the slope at that value of x is still going
to be the same.
The graph is just been moved vertically, up
or down.
So the slope at that given value of x is not
going to change.
So it makes sense to say that, or know, that
we have these three original functions but
yet the derivative of each of these functions
is 4x^3 for all of them.
That is because the derivative of a constant
is equal to zero.
So whether there is a plus zero, plus two,
or minus five, all of those are basically
going to disappear through finding the derivative
because the derivative of a constant is equal
to zero.
Thus you get the same answer three times.
So to match this definition of the antiderivative,
and remembering and realizing that we are
going to be given the derivatives and having
to go back and try and go back and find the
original function... or the antiderivative...
we are going to change the notation a little
bit.
We will say we are given the lower case f.
When we are given this derivative for the
definition...
When we look at the derivative, we are going
to say that this is a lower case f.
When we go through the indefinite integration
process we are going to be trying to find
the original function which we are going to
call capital F(x).
The definition of an antiderivative, a function
F is AN antiderivative of.... we are not going
to be able to find the actual original function...
we are not going to be able to find out what
that constant was at the end of the function
if there was one unless we have some more
information given... so is AN antiderivative
of lower case f on an interval I if the derivative
of capital F(x) is equal to lower case f(x)
for all x in I.
So this right here is basically telling you
how you can check your work.
When you are given something that you are
told is a derivative, and you go through the
indefinite integration process to find the
antiderivative, if you want to make sure your
work is correct just take the derivative of
your answer to see if you arrive at the original
derivative you are given.
The antiderivative of f(x) equals 4x^3...
this is kind of getting into the formal definition
I am standing in front of... the antiderivative
of f(x)=4x^3 is F(x)=x^4+c.
Ok.
The derivative of that constant is equal to
zero.
And again without extra information given
to us we cannot find out what that vertical
shift or that constant actually is at the
end of the function.
And this is with the plus 'c' is called the
family of all of the possible antiderivatives.
Now we are also going to be learning something
called a definite integral or work through
the process of definite integrations.
That is going to allow us to find the integral
with a particular boundary.
This is an indefinite integral, so there is
no boundary and thus we are going to be looking
for that original function.
But I probably should not have brought that
up because we are not going to deal definite
integrals in this video.
Representation of Antiderivatives.
If F is an antiderivative of lower case f
on an interval I, then capital G is an antiderivative
of f as well on the interval I if and only
if G is in the form of G(x)=F(x)+c.
So we are saying that we need to add that
plus c at the end of our answer to represent
the General Antiderivative or to represent
the family of all antiderivatives.
Let's get to that next page of notes.
Here we have the notation for antiderivatives.
dy/dx, the derivative of y with respect to
x is equal to f(x)... that lower case f again
indicating the derivative instead of the f
prime of x because the notation that was in
our definition of antiderivatives.
We are going to take this dx, or it could
be dt, it could be dtheta, whatever we found
the derivative of with respect TO that variable.
We are going to take that dx in this case
and multiply it to both sides of this equation
for convenience sake and say that dy is equal
to f(x) times dx.
You will see why that is down here in a second.
Now we are going to be asking what is y?
What is the original function if dy=f(x)dx?
In other words we want to go from the derivative
to the original function.
That means that we want to do an indefinite
integration.
So that indefinite integrations symbol, the
integration symbol here.... this sort of long
stretched out S. Think of it as the sum idea.
Because when we do definite integration we
are not there yet.
So we are going to have this indefinite integration
symbol here of f(x)dx.
Now f(x) is our integrand.
The dx is our variable of integration.
You know, sort of like the variable that we
found the derivative with respect to is now
going to be undone as the variable as the
variable of integration.
Then we have the equals capital F(x), that
original function again, plus c our constant
of integration.
That constant that we cannot find unless we
are given a little bit of extra information.
The integral of f(x)dx is read the antiderivative
of f with respect to x.
So the antiderivative of f(x) with respect
to x, the operation of finding all solutions
of this equation is called anti-differentiation
or indefinite integration.
I said those words again through the whole
video, but I wanted you to basically see that
concrete definition there.
Well, here we have a list of our basic differentiation
formulas.
We have the derivative of a constant equals
zero.
The derivative of a constant times x is equal
to just that constant.
So the derivative of 2x is equal to 2.
We have the derivative of a constant times
a function is equal to that constant times
the derivative of the function.
We have the derivative of the sum of terms
is just the derivative of each individual
term whether it is addition or subtraction
separating them.
That is fine.
And then we get to....
I am going to try and leave all these up here
as I do my examples.
So I kind of crammed these together.
I put a star here.
We have the biggie that we are going to be
dealing, the Power Rule.
Which says that the derivative with respect
to x of x to the n is equal to n times x^(n-1).
So when we do the derivative we reduce that
exponent by a value of 1 and take the original
exponent and move it down out front.
So say the derivative of x cubed is 3x^2,
at least the derivative with respect to x.
Now integration is just the opposite of...
the indefinite integration is just the opposite
of doing the derivative.
So the indefinite integral of...
Excuse the interruption there.
So we have the integral of 0 with respect
to x is equal to c.
Just the opposite of the derivative of C is
equal to zero.
We have the integral of k dx is equal to that
k times x plus C. Again, just the opposite
of finding the derivative.
We have the integral of a constant times a
function with respect to x is equal to the
constant... that same constant we can pull
that out in front of the integral notation...
and that is the constant times the integral
of that function.
We have the integral of two functions being
added together is, we can pull that apart
and give each terms its... or find the integral
of each individual term again with the plus
or minus in the middle.
It does not make any difference.
But again, the big deal here is the Power
Rule.
The integral of x to the n dx is equal to
x^(x+1), we are taking that exponent and raising
up by one as opposed to reducing it by one
as we did with the derivative process, and
then dividing that term by n+1.
So we are going to raise that exponent by
1 and then divide by that same exponent that
we have after we have increased it by a value
of 1.
Of course this is plus C, that constant at
the end.
Now we cannot divide by zero, so n cannot
equal negative one with this Power Rule for
integration.
Now the derivative of sine with respect to
x is cosine, so the integral of cosine dx
is equal to sine of x plus C. The derivative
of cosine is equal to negative sine, so the
integral of sine... now that negative has
kind of gone away but we have basically moved
it to the other side of the equation.
Multiply both sides by negative one if you
will.
So the integral of sine of x dx is equal to
negative cosine of x plus C.
We have the derivative of tangent with respect
to x is secant squared, so the integral of
secant squared x dx is equal to tangent of
x plus C.
We have the derivative with respect to x of
secant is secant x tangent x, so the integral
of secant x tangent x is equal to secant x
plus C. Cotangent gives us negative cosecant
squared with derivatives, so the integral
of cosecant squared x dx is equal to negative
cotangent of x plus C. And again the same
thing with the cosecant function which gives
us negative cosecant x times cotangent x when
we do derivatives.
The integral of cosecant x cotangent x dx
is equal to negative cosecant x plus C.
We are going to refer to all of these as we
go through our many examples I am about to
do with you.
And please be aware that we when you do the
integration process, you need to make your
problem exactly match what you see here in
these formulas.
So if we were trying to find the integral
of cosine squared, that is not exactly cosine
so that could potentially give us some problems
as we are just starting to learn our basic
integration formulas.
I have a lot of examples so let's get those
started.
Oh.
Excuse me.
let's get those examples started:)
Here is our first two examples.
We want to find the indefinite integral of
5xdx.
So we are going to use this rule right here
that says the integral of k times the function
dx is equal to that constant pulled out times
the integral of f(x)dx.
So we are going to take this constant of 5,
pull it out, and write this as 5 times the
integral of xdx.
Then we are going to use the Power Rule to
find the integral of x.
Now this has a power of 1, so we are going
to again use this Power Rule.
This is the biggie here.
We are going to raise that exponent by 1 and
then divide by that increased exponent.
So it is going to be 5 times x squared, so
we have raised that power of 1 by 1, so now
we are going to divide by that 2 plus C. Now
we are going to take that 5 that is in front
of this parenthesis here and distribute it
and get 5x^2 over 2 plus 5C.
Now C is defined as a constant... any constant
so writing 5C is a little bit... it is unnecessary
and over complicated.
We can just simply write that as C. So we
can rewrite this as 5x^2/2 plus again this
value of C.
This idea of taking this constant and pulling
it out front, finding the integral, and then
multiplying it back in is always going to
create this... basically over complicated
version of the constant.
So we are going to streamline that process
in this next example and show you that is
not really necessary to show you pulled the
constant from the function before you find
the integral.
And the four steps that we are going to with
all of these problems is we are going to look
at the original problem, we are going to rewrite
it to get to match one of these integration
formulas... these basic integration formulas
exactly, we are going to apply that integration
or integrate, and then simplify the expression
or whatever is left.
So when we see -8 over x cubed, this is a
quotient here... right.
We don't have a quotient rule over here.
We don't have a Product Property either.
We only have the Power Rule.
So I need to get this into some kind of format
that allows me to use that Power Rule.
We are going to do that by writing this so
that we don't see that fraction bar, so that
we don't see the quotient.
So we are going to write this as the integral
of negative eight x to the negative third
power dx.
So now I my division sign is gone and I don't
see it as a quotient.
It is just a term that I can apply the Power
Rule to.
So again, we are going to take this exponent,
we are going to increase it by 1, and -3 plus
1 is -2.
So this is going to become -8... notice that
I am just letting it just sort of float along...
I am not pulling it out in front of the integral
symbol first.
Just going to have the -8 float along, and
we have -8 times... take that exponent and
raise it by 1 so we have x^-2... and again
we are going to divide.
Again here we go, when we integrate x^n we
get x^(n+1) and we are dividing by n+1.
So we are going to take that exponent and
increase by one and then also divide by that
n+1 as well.
I just want to make sure you get that completely
set in.
This plus C. So -8 divided by -2 is 4x^-2
plus C. And we are going to write this so
we don't have a negative exponent and have
4 over x^2 plus C.
This is the answer to our original problem.
And I took up a lot of space here, plus my
next example is not on the board so I will
be right back.
For our next example we want to find the antiderivative
of 1 over x^2 times the square root of x.
Now again we have a fraction bar which means
quotient and we don't have any quotient rules
in our basic integration formulas.
We have the Power Rule.
So we are going to have to write this....
We also have of course a product going on
here and there is no produce rule either,
just the Power Rule.
So we need to rewrite this so that we can
use that Power Rule.
That means we need to get rid of the fraction
bar.
So this is going to become the integral of
x^-2 times x^-1/2.
I am going to write that radical as a power
over root, the index here is 2 so that is
x to the 1/2.
We move that out of the denominator and we
get x^-1/2... dx.
Ok.
So we got rid of the fraction bar.
But we still have the product thing going
on here.
So we need to take those like bases and multiply
them together and get the integral of x...
when we multiply like bases we add the exponents...
so this is -2 plus -1/2, or...
What would that be?
That would be -2/1 minus 1/2.
You need a common denominator of course, so
-4 minus 1 would be x^-5/2 dx.
Now I have the problem in a format that I
can apply the Power Rule.
That means that we are going to take this
power, increase it by 1, divide by that same
value, and so that becomes x to the -5/2 plus
2/2... that is of course 1 with the common
denominator written out.
Now -5/2 plus 2/2, that is -3/2 so we are
going to divide by -3/2... now plus C. So
we have x to the -3/2 power divided by -3/2
plus C. And when you divide, of course we
know that when we have this fraction in the
denominator we are dividing by a fraction.
That is the same as multiplying by the reciprocal.
So we are going to take this bottom value
of -3/2 and flip it up and make it a multiplication
of -2/3.
We get a final answer of -2/3 x^-3/2 plus
C. Now if you really want to get, you know,
do your simplification very throughly and
get rid of all your negative exponents this
can be rewritten as -2 over 3x^3/2 plus C.
And a lot...
I am sure that a lot of your teachers are
going to be perfectly fine with having this
fractional exponent.
You could however write this as -2 over 3
times x^1 times x^1/2 plus C. You see what
I am doing there.
When you multiply like bases you add the exponents.
And when you add 1 and 1/2 you get 3/2.
That basically...
You can think of this as 3/2 being x^2/2 times
x^1/2 and of course 2/2 cancels out to be
1.
And then we can put that radical notation
back in because x^1/2 is the square root of
x.
So you could write this as -2 over 3x times
the square root of x plus C. So you know,
it just depends on how far you need to go
in your simplification process to make your
teacher happy and of course you might need...
You might have a right answer, but not have
you answer in the format that is in the back
of the book.
So keep working on that simplification until
you get something that is in the back of your
book so you can check your work.
It is probably going to be in this format
here with the x^3/2 in the denominator.
Here we want to find the antiderivative of
4x-3x^2 with respect to x.
And this rule here says that if you want to
find the integral of two terms being added
or subtracted together, we just find the integral
or the antiderivative of each of those individual
terms and apply the plus or minus function
in the middle.
So, we ar going to have the antiderivative
of 4x is 4x taking that exponent and increase
it by 1 so it is 2 and we are going to divide
by 2 using that Power Rule again minus 3 x...
we are going to take that exponent and raise
it by 1 which gives us 3 and divide by 3 and
then apply that constant at the end.
For each of the terms you would have a constant
for each of those antiderivatives, but a constant
plus a constant is just another constant.
So we can write that plus C at the end just
once.
4/2 is 2x^2.
3/3 is 1 so we have minus x^3 plus C. That
is end of that example.
I have got two more, or three more examples.
So let's get those started.
So we want to find the antiderivative of x+4
over the square root of x dx.
Well, we got that fraction bar.
We have that quotient that we need to get
out of there.
We need to rewrite this again so we can apply
the Power Rule.
We are also going to want to get rid of that
radical symbol at the same time.
That will allow us to get the problem into
the format that we need.
So we are going to...
Let's see.
There is really nothing in the denominator
that we can flip up.
So what we are going to do is we are going
to see that we have 2 terms in the numerator
and we are going to pull those two terms apart
and write two separate fractions for now.
That is going to be the integral of x over...
now I am going to write this square root of
x back into that rational exponent form that
is easier to work with.
So we have a power of 1 and root of 2.
It is x to the 1/2 power plus 4 over x^1/2.
And again all with respect to x, we are doing
this integration here.
And now when you divide like bases you subtract
the exponents.
So we have an exponent of 1 and we have an
exponent of 1/2.
When we divided those like bases now in this
format, 1-1/2 is 1/2.
So we are going to have the integral of x^1/2
plus 4, and I want to take this x^1/2 out
of the denominator so we have 4x^-1/2 dx.
Now I have...
I can use the summation rule and of course
I am going to use the Power Rule on each of
those two terms.
So we get...
Let's see.
We are going to use the Power Rule on x^1/2.
You are going to increase that exponent by
1, so 1/2 plus 1 is 3/2.
We have x^3/2 divided by 3/2 plus 4 times
x to the ..... We are going to take this exponent
now.
Working on the second term.
Raise it by a value of 1, so -1/2 plus 1 is
1/2.
We are going to divide by that 1/2.
Now plus our overall constant at the end.
We are going to take each of these denominators
and flip them up.
So we have this 3/2 getting flipped up.
The same thing as divided by 3/2 is multiplying
by 2/3.
So we have 2/3 x^3/2 plus... now if we take
1/2 and flip that up, 4 times 2 is equal to
8x^1/2 plus C. That is our general antiderivative.
Ok.
So here we have x+1...
We want the antiderivative of x+1 times 2x+4
dx.
Well.
There is no product rule.
We only have the...
The most powerful rule we have at our disposal
is the Power Rule.
So we are going want to distribute these binomials
together and get this written as a polynomial
form, so we can use the Power Rule on each
of those individual terms.
So we have the integral of x times 2x is 2x^2...
x times 4 is 4x and 1 times 2x is 2x, so 4x+2x
make 6x... and 1 times 4 is equal to 4...
all with respect again to the dx.
So now we can do the integral or find the
antiderivative of each of those three terms.
We get... take this 2, so we have 2 times
and raise that exponent by 1 so we have x^3
and divide by 3... plus 6 times x and raise
that exponent by 1 and divide by the same
value plus 4x because the integral... the
antiderivative of a constant is equal to the
constant times x plus that overall constant
of C. And... um..
Not much to clean up here.
We have 2/3x^3 plus 6/2 is 3 so 3x^2 plus
4x plus C. Now if you happen to be using a
TI-89 or TI-NSPIRE CAS version to check these
problems... check you work by basically telling
it to do the integral for you... it will not
be adding the plus C at the end of the expressions
or answers so please be aware of that if you
are attempting to check your work with those
graphing calculators.
I have got a few more examples.
Here we go.
Last two.
Thank you for watching by the way.
I hope this is helpful.
We want to find the antiderivative of 3cos(x)-2sin(x)dx.
If you are comfortable with this and everything
is going really really well, you might be
able to just write the answer.
I am going to take the extra step to rewrite
this very specifically so that you see that
we are coming up and matching these integration
formulas... these basic integration formulas
for the trig functions exactly before applying
them.
So we are going to rewrite this as....
Let's see...
We are going to use the sum and difference
property to separate those integrals.
We are also going to move the constants out
front.
So this is equal to 3 times time the integral...
or we want to find the antiderivative or the
integral of cos(x)dx minus 2 times the integral
of sin(x)dx.
Each one of these needs the dx.
We have to remember what we are doing.
The integration with respect to x.
So the integral of the cosine of x is equal
to sin(x)+c.
So this is going to be 3 times sin(x).
Because I have all of these multiple terms
I am going to leave the plus C at the end.
Now minus 2 times the integral of sine is
negative cosine x, so -2 times -cos(x) plus
C. Now what... not much left to clean up.
We just have -2 times -1, so we have 3sin(x)+2cos(x)
and again don't forget that plus C at the
end so we have an answer for the family of
all of the antiderivatives.
Now we want to find the antiderivative of
the sin(x) over 1 minus sin(x) squared.
We have already discussed how there is no
quotient rule in our list of general integration
formulas.
So if we cannot get this to exactly match
something we have over here, or that we can
separate it into separate terms we are going
to have a problem.
We also don't have anything, any of these
formulas that says specifically 1-sin^2(x)
and that could be an issue.
So what we are going to do to attack this
is to play around with our trig identities
and see if we can't get this rewritten to
a more friendly format.
So 1-sin^2(x).
If you remember your first Pythagorean Identity,
sin^2(x)+cos^2(x) equals 1.
So if you subtract that identity... both sides
by sin^2(x) you get 1-sin^2(x) equals cos^2(x).
So this becomes the integral of sin(x)/cos^2(x)
dx.
Well I still have a fraction in there.
It is still a quotient so that is not good.
I also don't have...
This is the integral of cosine, sine, secant
squared, secant... you know there is nothing
in here.
I don't have a formula for doing the integral
of sin(x)/cos^2(x).
So what to do?
Well if I write this as the integral of sin(x)/cos(x)
times 1/cox(x) dx.
Now I can write sin(x)/cos(x) as tan(x).
So we have the integral of tan(x)...
1/cos(x) is sec(x) dx.
And um... hmm.
The integral, see I have gotten it rewritten
to match exactly one of these formulas here.
The integral of sec(x)tan(x)dx is equal to
sec(x)+C. So this is equal to sec(x)+C. I
have got that problem.
That is the end of my last example.
SO!
I am Mr. Tarrou.
Go Do Your Homework:D