We want to find the arc length
of y=2x squared - 3 on the
closed interval from one to two.
Looking at the graph of the function,
notice when X is one, or at this point
in the function, 1,5, and when X is two,
we're at this point in the function, 2,11.
So when we find the arc length
on this closed interval,
we're finding the length
of this piece of the curve.
Since we have a function of
X, we'll find the arc length
using this integral formula
here, where the arc length,
S, is equal to the
integral of the square root
of the quantity 1 + F prime of X squared,
integrated with respect to X, from A to B.
So our function F [X] = 2x squared - 3,
and F prime of X, the derivative function,
would be equal to 4x.
Therefore, the arc length,
S, is equal to the integral
of the square root of 1
+ F prime of X squared,
which would be 4x squared,
integrated with respect to X,
from A to B, or in this case, one to two.
Notice how this gives us the integral
of the square root of 1 + 16x squared...
There's no simple integration
technique to integrate this,
so we'll first evaluate this
on the graphing calculator,
and then we'll evaluate this by hand
using a formula from integration table.
Using the graphing calculator,
we would press Math,
arrow down to function integration,
which is option nine, press Enter.
Enter the integrand,
which is the square root,
so second X squared of 1 + 16x
squared, closed parenthesis
for the square root, comma,
the variable of integration,
which is X, comma, lower
limit of integration is one,
comma, upper limit of integration is two,
closed parenthesis and Enter.
So the approximate arc
length would be 6.0859.
Now, let's evaluate this another way
using a formula from an integration table.
Notice how the integral fits
the form of this formula here,
but let's rewrite 16 x squared
as four X to the second,
and let's also write this as
four X to the second plus one.
So to make this fit the formula better,
we can rewrite this as the
integral from one to two
of the square root of
4x to the second + 1.
Notice how, in this form, it's easier
to recognize that U is equal to 4x...
And A is equal to one.
But notice if U = 4x, than differential
U would be equal to 4dx.
But notice if we solve this 4dx....
DX = 1/4 DU, 'cause if we
write this in terms of U,
we'll have an extra factor of 1/4.
Let's go ahead and do that.
This is equal to the integral of,
notice when X is one, U is
equal to four times one or four,
and when X is two, U is equal
to four times two or eight.
Now we'd have the square root of,
again, we have U=4x, so
we do have U squared plus,
again, A=1, so we have A squared,
but DX is equal to 1/4
DU, so we have a DU here.
Let's factor out the 1/4,
and now I can go ahead and
apply the integration formula.
We'll have 1/4, and our
antiderivative is given here,
so I'll have another factor of 1/2.
Let's go ahead and factor that out.
So we have times 1/2, then
we have our antiderivative...
We'll evaluate this at eight, then four,
and then find the difference.
So we have 1/8...
So first, U is equal to eight,
and again, we already know A is one,
so we'd have eight times the square root
of eight squared plus one
squared, that's going to be 65,
plus, again, A squared is
one, so we have natural log.
We don't need the absolute
value because notice
how U and A are both positive.
So we have natural log
of U, which is eight,
plus the square root of
U squared plus A squared,
that's eight squared
plus one squared or 65...
Minus, when U is four, we have
four times the square root
of four squared times one
squared, that's 17 plus,
again, A squared is one,
so we have natural log
of U, which is four, plus the square root
of four squared plus one squared or 17.
Now, let's go to the calculator.
It might be easier to
evaluate this in pieces,
I'm going to go ahead and
try to do it all in one step.
In parenthesis, we have
1/8, and then we'll have
two open parenthesis here.
Eight square root 65, closed
parenthesis for the square
root, plus natural log
eight plus square root 65.
Here we need a closed
parenthesis for the square root,
then for the natural log, and then
for this first expression, and then minus,
open parenthesis, four square
root 17, closed parenthesis.
Then, we have plus natural log...
Of four plus square root 17.
Here we have a closed
parenthesis for the square root,
then for the natural log, then
for the second expression,
and then finally, one more for
the outer bracket, and Enter.
Now notice how we do have the same value.
Approximately 6.0859 for the arc length.
I hope you found this helpful.
