In this video,
I'd like to address a common question
when evaluating a def integral using substitution.
And the question is, when evaluating a def integral
using substitution, do we need to determine new limits
of integration?
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, we first need to decide
if we want to let u equal sine x or u equal cosine x.
And because the derivative of cosine x is equal
to negative sine x, let's let u equal cosine x.
So, if u equals cosine x,
the next step is to find differential u.
Differential u is equal to negative sine x times dx.
Now, analyzing the integral, because u equals cosine x,
we can substitute u for cosine x.
Notice how this leaves negative sine x, dx,
and this matches perfectly,
because we know differential u equals negative sine x, dx.
So, writing this in terms of u,
we would have the integral of, again, u equals cosine x,
so we'd have u, and then negative sine x, dx, equals du.
So we'd just have the integral of u, du.
And now let's talk about the limits of integration.
The limits of integration from pi over two to pi
are for x, not u, so we can go ahead and leave them
in terms of x if we wish.
Let's just make a note here.
These limits are for x, not u.
If we leave these limits of integration in terms of x,
once we find the antiderivative in terms of u,
we need to rewrite the antiderivative in terms of x
before evaluating.
The other option would be to find new limits of integration
for u, and let's show this below.
We'd have the integral of u, du.
To find the limits of integration for u,
we would substitute these x values into our equation for u.
So, notice when x equals pi over two,
u is equal to cosine pi over two, and so for a quick review,
cosine pi over two is equal to zero.
So, the lower limit of integration for u would be zero.
And then when x is pi, we'd have u equals cosine pi,
and cosine pi is equal to negative one.
So, if we use these limits of integration,
we'd use the antiderivative in terms of u,
and if we leave limits of integration in terms of x,
we'd use the antiderivative in terms of x.
So, the next step will find the antiderivative in terms of u
and because this is u to the first,
we would just have u to the second divided by two,
and we'll write this in terms of x, so we'd have 1/2,
and then u squared is equal to cosine x squared
or cosine squared x.
Now that we have the antiderivative back in terms of x,
we can evaluate this using our limits of integration
in terms of x.
So, we'd have 1/2 times...
When x is pi, we'd have cosine squared pi.
And when x is pi over two,
we'd have cosine squared pi over two.
We know that cosine pi is equal to negative one,
but we'd have negative one squared,
and cosine pi over two is equal to zero,
so we'd have zero squared.
Notice how this simplifies to 1/2 times one,
which equals 1/2.
Looking at the same example below,
we know the antiderivative
in terms of u is u squared divided by two.
Let's go ahead and factor out the 1/2.
But because we have the limits of integration in terms of u,
we can evaluate this in this form.
We'd have 1/2 times...
When u equals negative one, we'd have negative one squared,
and when u is zero, we'd have zero squared,
giving us the same result of positive 1/2.
I think it's important to be aware
of both of these techniques because,
depending on the source, it could be shown either way.
Now, let's also verify this on the graphing calculator.
Let's first evaluate the integral in terms of x,
and then we'll evaluate it in terms of u.
Let's first verify we are in radian mode,
so press the mode key.
Notice how radian is highlighted,
so we'll go back to the home screen.
Press math and then option nine for the def integral.
We're integrating from pi over two,
so second caret brings up the pi divided by two,
right arrow to pi,
so second at caret,
right arrow.
Our function is negative cosine x,
close parenthesis, sine x,
close parenthesis, right arrow,
and then x for dx and enter.
Verifying 1/2 is correct.
Let's also show this in terms of u,
so we'll enter this integral here,
so we'll press math,
nine,
integrating from zero, so zero, right arrow,
to negative one,
right arrow,
and then the integrating function here is just u,
so alpha five,
right arrow, and then u again for differential u,
so alpha five, and enter.
And of course we get the same result.
I hope you found this comparison helpful.