This lesson will show how to evaluate
the given definite integral
using the method of U-substitution.
The first step is to recognize that U-substitution
is required because the integral function
is the sign of one half X, not just the sign of X.
So here U is equal to the inner function of one half X
and therefore differential U is equal to
the derivative of one half X times DX,
which gives us one half DX.
And now looking at the integral
and notice how we don't have one half DX,
and therefore we will solve for DX
by multiplying both sides by two.
Two times DU is equal to two times one half DX
giving us two DU equals DX.
So now we know we can substitute U for one half X
and we can substitute two DU for DX.
And we need to remember right now,
the limits integration are for x not U.
This is X equals zero, and this is X equals two PI.
So now let's write this in terms of U.
DX is equal to two DU, we'll factor out the two,
and sin one half X is now sin U.
And now we need to make a choice.
If we want to leave this in terms of U to evaluate,
we need to determine the limits
of integration for U using the equation U equals one half X
or we can find the anti derivative with respect to X
and use the original limits of integration.
In this video, I will show both methods.
So I'm going to go down here and write two times integral
of sign U DU, and we'll come back to this shortly.
And now I'll show how to find the limits
of integration for U.
When X is equal to zero,
U is equal to one half times zero or zero.
And when X is equal to two PI,
U is equal to one half times two PI or pie.
And now we can evaluate this, leaving this in terms of U.
The anti derivative of sin U is equal
to negative cosign U plus C.
We leave the plus C off when evaluating definite integrals.
So the anti derivative is two times negative cosign U
or negative two cosign U.
Let's go ahead and factor out the negative two.
And now we determine negative two times the difference
of big F of pie and big F of zero
where big F of pie is cosign pie
minus big F of zero is cosign zero.
This gives us negative two times the quantity.
Cosign pie is negative one minus cosign zero is one.
Here we have negative two times negative two
which is positive four.
The definite integral is equal to positive four.
But again, for this example, I also want to show
how we could have determined the anti derivative
in terms of X and use the limits integration
from zero to two pie.
So if we go back to the indefinite integral written
in terms of U, two times the integral of sign U DU,
we already know from above
is equal to negative two cosign U plus C
which means the anti derivative in terms of X
would be negative two cosign of one half X.
and here because we have the anti derivative
with respect to X, we would evaluate this
using the original limits of integration,
which are from zero to two pie.
And let's just show we get the same result.
Factoring out the negative two first.
We would have negative two times the quantity cosign
of one half times two pie minus cosign
of one half times zero
which gives us negative two times the quantity.
Well, here we have the cosign of pie,
which we know is negative one minus cosign zero,
which is one.
And of course we get the same result.
Negative two times negative two, which is positive four.
And I do like showing the second method occasionally
because sometimes textbooks won't show the U substitution.
They'll show the anti derivative only in terms of X.
And finally, before we go take a look at the graph
of Y equals a sign of one half X
over the close interval from zero to two pie
and shade the area bounded by the function of the X axis.
We have this purple area here
and because the area is above the X axis
the value we found is equal to this shaded area.
I hope you found this helpful.