[Instructor] Welcome to a lesson on indefinite integration
or anti-differentiation using the substitution method.
The substitution method is a method used
to find the integral of a composite function.
The substitution method simplifies integral so that we can
find the antiderivative using a basic integration formula.
So notice how on this integral we have a composite function
where the inner function is g of x and notice
how the integrand also contains g prime of x using
substitution we can simplify this to the integral of
f of u inequated with respect to u.
Notice how by performing substitution, we'll make a change
of variables.
And here the steps will follow.
Step one we'll choose a substitution u equals g of x.
Usually it is best to choose the inner part of the composite
function such as a quantity raised to a power.
Two, we'll compute differential u where differential u
equals g prime of x times dx.
Three, write the integral in terms of the variable u.
Four, find the resulting integral in terms of u.
Five, replace u with g of x to obtain an antiderivative
in terms of x.
And step six, we can always check our answer
by differentiating.
Let's look at our first example.
Notice how we have the quantity two x to the fourth plus
seven raised to the seventh power so the general strategy is
to let the quantity raised to the power be equal to u.
So here we'd let u equal two x to the fourth plus seven.
Then part of the substitution method is to recognize that if
we let u equal part of the integrand, then its derivative
will also be in the integrand.
Notice how the derivative of two x to the fourth plus seven
is going to be eight x to the third.
So once we select u, the next step is to find differential
of u, where differential of u will be equal
to the derivative of two x to the fourth plus seven
with respect to x times dx, so we'd have eight x
to the third times dx.
Now looking at the integral again, since u is equal
to two x to the fourth plus seven, we'd have u
to the seventh here.
Notice the remaining part of the integral is eight x
to the third dx which is equal to differential u.
We don't always have a perfect match here
for differential u, sometimes we have to manipulate this
equation here, but in this case we have a perfect match.
The next step is to write the integral in terms of u,
so in terms of u we'd have the integral of again,
this is u to the seventh.
And then eight x to the third dx is equal to du.
Notice how this integral simplifies nicely to the integral
of u to the seventh, integrated with respect to u.
So in this form we can apply the basic power rule
of integral given here.
So we would have u to the seventh plus one, that's eight,
divided by eight plus c.
Now if we want 1/8 u to the eighth plus c.
But our goal here is to find the antiderivative
in terms of x not u.
So now we substitute two x to the fourth plus seven for u.
So the antiderivative would be 1/8 times the quantity two x
to the fourth plus seven raised to the eighth plus c.
So this is our antiderivative, which means big F of x is
equal to 1/8 times the quantity two x to the fourth plus
seven raised to the eighth plus c, and the derivative
of big F of x should give the integrand function eight x
to the third times the quantity two x to the fourth
plus seven raised to the seventh.
Let's go and check this.
Big F prime of x is equal to 1/8 times,
to find the derivative of two x to the fourth plus seven
raised to the eighth, we'd multiply by eight
and then we'd have two x to the fourth plus seven
to the eighth minus 1/7 power and then applying
the Chain Rule, we multiply by the derivative of two x
to the fourth plus seven which is eight x to the third.
And the derivative of a constant of zero, so simplifying.
Notice here the eighths simplify out, so we'd be left
with big F prime of x equals eight x to the third
times the quantity two x to the fourth plus seven
raised to the seventh, which notice is the integrand
function f of x which means our work is correct.
Let's look at another example.
This first is using rational exponents, because the index
is two and we have the quantity four plus x squared
to the first, we can write this as the integral of x
divided by the quantity four plus x squared
to the one half dx.
And again because we have four plus x squared raised
to the one half, we'll let u equal four plus x squared.
And now we'll find differential u, differential u is equal
to the derivative of four plus x squared with respect
to x times dx.
So that would be two x times dx.
And looking at our integral again, we know four plus x
squared is equal to u, notice how we're left with x dx
and over here we have du equals two x dx.
Let's go ahead and solve this equation for x dx
by dividing both sides by two.
So simplifying we have one half differential u is equal
to x dx.
And now I'll part this in terms of u.
So we'd have the integral of, notice how our denominator
is going to be u to the one half, and then we have x dx,
which we know is equal to one half du.
So let's factor out the one half and then we have our
differential u and therefore our denominator is going
to be one.
Before we apply our integration formula though, we do
want to move this up to the numerator which will change
the sign of the exponent.
So we can write this as one half times integral of u
to the negative one half, differential u.
And then we'll find the antiderivative in terms of u.
So we'd have one half times u to the negative one half
plus one, that's u to the one half
divided by one half plus c.
Simplifying it again we have one half, and then dividing
by one half is the same as multiplying by two,
so times two over one, times u to the one half plus c.
Simplifying.
We have u to the one half plus c.
And now we substitute four plus x squared for u.
So the antiderivative in terms of u is going to be
the quantity four plus x squared raised to the one half
plus c, or if we want we could write this as the square root
of four plus x squared plus c.
Either of these two forms of the antiderivative would
be acceptable.
Again, let's go ahead and check this.
And to check this we now know that big F of x is equal to,
let's use this form here, so we can find the derivative,
the quantity four plus x squared, to the one half plus c.
So big F prime of x should give us the integrand function,
so we've multiplied by one half, one half times the quantity
four plus x squared, and then one half minus one
is negative one half times the derivative of the inner
function which would be two x.
And the derivative of our constant is zero, so simplifying
the two's simplify out, we have a negative exponent here,
so we'll move to the denominator, so this is equal to x
divided by the quantity four plus x squared to the positive
one half or if we want, x divided by the square root of four
plus six squared.
So once again this verifies our antiderivative is correct.
Let's look at one more example.
Notice here we have an exponential with an exponent
of negative two x to the second, so notice how the exponent
is degree two and we have a degree one term here,
so we'll let u equal negative two x to the second.
And now we'll find differential u, differential u
is equal to negative four x dx.
Looking at our integral we don't have a negative four x dx
but we do have an x dx so let's solve for x dx
by dividing both sides by a negative four.
So we know negative one fourth du equals x dx.
We do have this coefficient of negative three here,
which we'll just factor out.
So let's write this in terms of u.
Again we'll factor out the negative three
and then we have the integral of, because negative two x
squared is equal to u, we have e to the u and we already
factored out the negative three, so we have x dx remaining,
x dx is equal to negative one fourth du, so we'll factor
out the negative one fourth.
And we have our differential u.
Simplifying again we have positive 3/4 times integral of e
to the u du, which is equal to 3/4 e to the u
is negative two x squared and then plus c.
So this is our antiderivative, which means big F of x
is equal to 3/4 e to the negative two x squared plus c.
So if we find big F prime of x, we should get the integrand
function negative three x items e
to the negative two x squared.
So our derivative is going to be 3/4 times the derivative
of e to the negative two x squared, which will be e
to the negative two x squared times the derivative
of negative two x squared that's negative four x.
The derivative of the constant is zero.
So now simplifying, we have a common factor of four
here and here.
So we have big F prime of x equals three times, this would
be negative x, so negative three x, e to the negative two x
squared which again is the integrand function f of x.
So once again, our work is correct.
Think we'll stop here for this part, in part two we'll take
a look at using substitution to evaluate definite integrals.
I hope you found this helpful.