In this video, I'd like to answer
a common question when evaluating def integrals
using the substitution method.
And the question is,
when evaluating a def integral using substitution,
do we need to determine new limits of integration?
And the answer is, it depends.
If we use the antiderivative in terms of x,
we do not need to determine new limits of integration.
However, if we use the antiderivative in terms of u,
we do need to determine new limits of integration for u.
Let's look at an example and we'll show both techniques.
So analyzing the integral, notice how
because we have the quantity
three x squared plus one raised to the second,
where three x squared plus one is the inner function,
let u equal three x squared plus one.
The next step we find differential u,
where differential u is equal to the derivative
of three x square plus one
with respect to x times dx,
so du equals six x dx.
And now let's analyze the integral.
Because u is equal to three x squared plus one,
we can write this part as u squared.
And then notice how we're left with four x dx
and we have differential u equals six x dx.
Let's focus on x dx since we can factor the four out.
Notice how if we divide both sides by six here,
we know that 1/6 du is equal to x dx,
which means we can substitute 1/6 du for x dx.
Now let's write this integral in terms of u.
So we'll factor out the four
and then we have the integral of,
this would be u squared and then x dx is equal to 1/6 du,
so we'll factor out the 1/6
and then we have our differential u.
And now let's discuss the limits of integration.
The limits of integration from zero to two are for x
so we could write down here x equals zero to x equals two
to remind us these limits of integration
are for x not u.
If we do this, we can find the antiderivative in terms of u,
write it back in terms of x
and then evaluate using these limits of integration.
The other option would be to find
new limits of integration in terms of u.
And let's show this below.
So we'd have four times 1/6, times the integral,
and then we have u squared du.
So notice how when x is zero using our equation for u,
you'd have u equals three times zero squared plus one,
so u would be one, which we can use as our
lower limit of integration in terms of u.
And then when x is two,
we'd have u equals three times two squared plus one,
which is equal to 13, so we can use 13
as our upper limit of integration for u.
So if we take this approach, we would use the antiderivative
in terms of u to evaluate the integral
and as I mentioned before above,
if we use limits of integration in terms of x,
we must use the antiderivative in terms of x
to evaluate the integral.
Now for the next step, we have four times 1/6,
that's 4/6 or 2/3 and now we find
the antiderivative in terms of u,
which would be u to the third, divided by three.
Limits of integration in terms of x are from zero to two.
Because the limits of integration are in terms of x,
we'll now write the antiderivative back in terms of x.
So here we'd have 2/3 times 1/3, that's 2/9.
And then u to the third in terms of x
would be the quantity three x squared plus one, to the third
and now we'll evaluate this.
We have 2/9 times a quantity when x is two,
three x squared plus one would be 13,
so we'd have 13 to the third.
And then minus when x is zero,
we'd have one to the third.
Let's evaluate this on the calculator.
We have two divided by nine,
times the quantity 13 cubed,
right arrow, minus one cubed,
right arrow, close parenthesis and enter,
which gives us 488.
Now looking at the example below
where we have the limits of integration in terms of u,
we still have 4/6 or 2/3,
we know the antiderivative in terms of u
is u to the third divided by three.
So simplifying we can write this as 2/9 times,
and again we have u to the third.
And because our limits of integration are in terms of u,
we can evaluate the antiderivative in this form here.
So we'd have 2/9 times, when u equals 13 we'd have 13 cubed,
minus when u is one we'd have one cubed,
and notice how this expression here
is the same expression we had above,
which we know is equal to 488.
Let's go ahead and verify this on the graphing calculator.
So we'll press math and then option nine.
Let's first verify the integral in terms of x.
So we integrate from zero to two,
so zero, right arrow, two, right arrow,
our integrant function is
four x times the quantity three x squared plus one,
raised to the second,
right arrow, and then x for dx, enter,
which verifies our answer is correct.
Let's also integrate in terms of u,
using this integral here.
So we'd have 2/3 or two divided by three,
and then math nine, integrating from one to 13,
so one, right arrow, 13, right arrow,
and then we have u to the second,
so for u we press alpha five and then squared,
right arrow, we enter u again for du.
So alpha five and enter,
and of course we get the same result.
So of course whichever technique is used,
the result is the same.
I hope you found this comparison helpful.