Assume that sine x equals
this Maclaurin series for all x.
We want to use the first two terms
of the Maclaurin series for sine two x squared
to approximate the integral.
So, because we're forming a polynomial
from the Maclaurin series,
we could say that we're going to use
a Maclaurin polynomial to approximate this integral.
But in this example,
instead of using this formula here for the Maclaurin series
we'll assume that we know the power series
for sine x centered at zero is this power series here.
And that way, to find the Maclaurin series
for sine two x squared
we can substitute two x squared
for x in the power series.
So, we can say that sine
two x squared would be equal to,
instead of x, we would have two x squared
minus, instead of x cubed divided by three factorial
we'd have two x squared cubed
divided by three factorial
and notice here
we have the first two non-zero terms
of the Maclaurin series,
and therefore, we would use these two terms
to approximate this integral.
And also notice how this second term is degree six.
So, if we did use this formula here
notice how this tells us we'd have to find
up to the sixth derivative of f of x equals
sine two x squared in order to find
these first two non-zero terms.
Let's go and find one more term.
We'd have plus, instead of x to the fifth
we'd have two x squared
to the fifth
divided by five factorial
and so on.
Let's go ahead and simplify these terms.
So, we'd have two x squared
minus,
here we have two x squared to the third
that would be eight x to the sixth
divided by three factorial which is six.
Let's simplify this to 4/3 x to the sixth.
So, we have minus 4/3
x to the sixth
and the next term
we have plus two x squared to the fifth.
That would be 32 x to the tenth
divided by five factorial
which is 120.
This simplifies to 4/15
x to the tenth.
And of course, for the Maclaurin series
this continues indefinitely.
But again, because we're using
only the first two terms
to approximate this integral
we could say the Maclaurin polynomial
the degree six
Maclaurin polynomial is equal to
two x to the second
minus 4/3 x to the sixth.
So, I'll use this polynomial
to approximate this integral.
So, this def integral
should be approximately equal to the integral
from zero to zero point seven six
of the Maclaurin polynomial
two x squared minus 4/3 x to the sixth
integrated with respect to x.
Let's evaluate this on the next slide.
So, I'll first find the anti-derivative
that would be two
times x to the third divided by three
minus 4/3 times x to the seventh
divided by seven.
So, we'd have 2/3 x cubed
minus four divided by 21 x to the seventh.
So, when x is 0.76
we'd have 2/3 times
0.76 to the third
minus four over 21
times 0.76 to the seventh
and then when x is zero,
notice both terms would be zero.
So, now we go to the calculator.
So, we have 2/3
times .76
to the third
right arrow
and then minus four divided by 21
times .76
to the seventh
right arrow
I think I did enter this parenthesis here
so, go ahead and enter one more on the right.
Of course, we have minus zero which we can leave off.
So, enter.
So, we have approximately 0.264755.
So, while this is the exact value of this def integral
it's our approximation for the given integral
of sine two x squared.
Now, before we go
let's compare the graph of these two functions.
The blue graph is the graph of
y equals sine two x squared
and the red graph is the graph of
the Maclaurin polynomial that we used
to approximate this integral.
Notice both functions are non-negative
over the interval from zero to 0.76
which I graphed here in purple
and therefore, the def integrals
are equal to the area under the functions
bounded by the x-axis.
So, we found the area under this red polynomial
and above the x-axis over this interval
to approximate the area under the blue function
on the same interval above the x-axis.
And looking at the area of the shaded region
we can tell our approximation is very good.
I hope you found this helpful.