Welcome to introduction
to antiderivatives and indefinite integration.
Let's first consider these two questions here.
If f prime of x equals one, then what would f of x be?
So we're given the derivative of f is equal to one,
the question is what would be the function of f of x.
So we're looking for a function that has
a derivative equal to one
so we should recognize the derivative of x
with respect to x is equal to one,
so our function f of x could just be f of x equals x,
but because the derivative of a constant
is also zero, we could add a constant on here
and the derivative would still be equal to one
so f of x equals x plus C.
Notice out here there's an infinite number of functions
that have a derivative equal to one.
In this case we said that f of x is the antiderivative
of f prime of x because its derivative is equal to one.
Looking at our notes below, an antiderivative
of a function f of x is any function F of x where
F prime of x equals f of x.
So normally when talking about antiderivatives,
we don't use this notation here.
We let the derivative function be equal to little f,
so in this case little f of x would be equal to one
and we let the antiderivative be equal to F of x,
so in this case, F of x would be equal to x plus C
and the reason F is the antiderivative of f
is because F prime of x is equal to one
which equals f of x.
Next, if f prime of x equals 2x, then what would f of x be?
So here we're given the derivative is equal to 2x,
we want to find a function that has the given derivative.
We should recognize the derivative of x squared
is equal to 2x, so we could let f of x be equal
to x squared but again because the derivative
of any constant is equal to zero,
we could add any constant here at the end,
so let's write plus C where C is any constant,
so f of x is the antiderivative of f prime
of x, and again we don't normally use this notation
when referring to antiderivatives.
We let the derivative function be f of x
and the antiderivative be F of x.
So here F of x is x squared plus C.
Again, the reason F is the derivative of f
is because F prime of x equals 2x which equals f of x.
So referring back to our notes again,
the antiderivative of a function f of x
is a whole family of functions written F of x plus C
which we've done here in black
where F prime of x equals f of x
and C represents any constant.
The antiderivative is also called the indefinite integral.
The notation used for the antiderivative
is shown here, where this first symbol here
is the integral sign and the dx on the end here
must be included: It indicates which variable
we're integrating with respect to,
and the function f of x is called the integrand.
So we say we antidifferentiate or integrate
or find the indefinite integral of a function.
This process is called antidifferentiation or integration.
And just like we have derivative formulas,
we also have integral formulas,
so here on the left we have common differentiation formulas
and here on the right we have common integration formulas.
Let's look at some examples.
Here we want to find the antiderivative
of x to the third minus 10x plus three with respect to x.
So one of the properties of integrals
that we'll often use is we can write this
as the difference of three separate integrals.
We can write this as the integral of x to the third
minus the integral of 10x plus
the integral of three.
Another property we'll often use
is if we have a coefficient like we have here
we can factor it out so we can also write this as
the integral of x to the third minus 10
times the integral of x plus the integral of three.
Now we integrate each of these separately.
So we integrate x to the third with respect to x,
we'll apply the power property of integration
shown here, so we add one of the exponent
and divide by the new exponent
and we'll leave the C until the very end.
So the antiderivative of x to the third
would be x to the power of three plus one or four
divided by four and we have minus 10 times
the integral of x, which is x to the first,
so we add one to the exponent and then divide
by the new exponent so we have x to the power
of one plus one which is two divided by two
plus the antiderivative of any constant
is the constant times x so the antiderivative
of three with respect to x would be 3x
and then we would have plus C, our constant of integration.
Now let's go ahead and simplify.
We write this first term as one-fourth x to the fourth.
Notice here we can simplify, there's one two in two,
and five twos in 10, so we have minus 5x squared
plus 3x plus C.
So this is the antiderivative of x to the third
minus 10x plus three, so we could say that
F of x is equal to one-fourth x to the fourth
minus 5x to the second plus 3x plus C.
This is the antiderivative function
because F prime of x equals
the integrand function of x to the third
minus 10x plus three, so whenever
we find an indefinite integral, we find an antiderivative,
we can always verify it by determining the derivative
and making sure it matches the integrand function.
Now for this example, we're not going to write this
as three separate integrals, we'll just integrate each term
but we do have to change the form of the second
and third terms in the integrand function
in order to apply the power rule
of integration shown here.
We'll write the square root of x using a rational
exponent, we'll write five over x to the second
using a negative exponent.
So let's write this as integral of 4x to the seventh
minus eight and now for the square root of x,
the index is two, the exponent is one,
so the square root of x is equal to x to the one-half,
and then we have plus five divided by x to the second,
is the same as 5x raised to the power of negative two,
so plus 5x to the power of negative two
integrated with respect to x, and now we'll just integrate
each term so the antiderivative of four x to the seventh
would be four times the antiderivative of x to the seventh
so we add one to the exponent, that'd be x to the eighth,
divide by the new exponent of eight,
minus eight times the antiderivative of x to the one-half,
so we'd have x to the power of one-half plus one,
that's one and a half or three halves,
divided by three halves, and then plus five
times x to the power of negative two plus one,
that's negative one divided by negative one,
and then don't forget the constant
of integration, so plus C.
Now we simplify, we notice four and eight simplify,
there's one four in four and two fours in eight,
so I can write this as x to the eighth
divided by two or one-half x to the eighth.
Remember our fraction bar means division
so this is minus eight and then divided by three halves
is the same as times the reciprocal of two-thirds.
We'll simplify this product in the next step.
And then here five divided by negative one
is negative five so instead of plus negative,
we'll just write minus five x to the power
of negative one plus C.
Let's continue simplifying, so we have one-half x
to the eighth, here nothing simplifies,
so we have minus 16 thirds x to the three halves.
Let's write this as minus five divided by x plus C.
So this would be your antiderivative
meaning the derivative of this function
would be equal to the integrand function
of four x to the seventh minus eight times
the square root of x plus five divided by x squared.
Next we have the integral of 3.5,
and notice how we have this dy here,
which means we're integrating with respect to y now,
not x, so we know the integral of k with respect
to x is equal to kx plus C, but because we're integrating
with respect to y not x, the antiderivative function
is equal to 3.5y plus C.
If we differentiate 3.5y plus C
with respect to y, we do get 3.5.
Next we have the integral of e raised
to the power of p integrated with respect to p.
Looking at the integral formula here on the right,
the integral of e to the x with respect to x
is equal to e to the x plus C
and therefore the integral of e to the p
with respect to p would just be e to the p plus C.
Let's take a look at two more.
Here we have the integral of five divided by x
integrated with respect to x.
In this case, it might be helpful to factor out the five
and write this as five times the integral
of one divided by x dx, and now looking
at our formulas, we see the antiderivative
of one over x or the integral of one over x
with respect to x is equal to natural log
of the absolute value of x plus C
so here we'd have five times natural log absolute value
of x plus C.
This last example looks a little bit different
because we have a product.
Let's first multiply this out and then we'll integrate,
so we'll have four products.
So we'll have the integral of x squared
and then we have minus three x, plus three x,
that's zero, and then we have three times negative three,
that's negative nine or minus nine.
And now we'll integrate each term,
so the integral of x squared with respect to x,
we have x to the power of, add one to the exponent,
that'd be two plus one or three,
divided by the new exponent, minus the antiderivative
of nine with respect to x would just be nine x,
and then plus the constant of integration.
Let's go and rewrite this as one-third x to the third
minus nine x plus C.
Okay, I hope you found this introduction helpful.