Welcome to an example of numerical integration.
We want to use the Midpoint Rule with n equals four
to approximate the definite integral.
And since n equals four, we'll be using four rectangles
to approximate this definite integral,
and because we're using the Midpoint Rule,
we'll use the x value that will be
the midpoint of each interval
to then determine the function value,
which will give us the height of each rectangle.
To illustrate this, let's consider this function here,
where the interval from zero to two
is divided into four equal intervals.
If we consider each interval,
notice how we're using the midpoint
of each interval to determine the function value,
which gives us the height of each rectangle.
So to find the area of each rectangle,
we'll find f of the midpoint,
and then multiply it by the width
of each rectangle, which will be delta x.
Now just to compare, if we use the left sum,
or the left side of each interval, to determine
the height of each rectangle, it would look like this.
If we decided to use the right sum, or the right side
of each interval to determine the rectangle height,
it would look like this.
But again, for our example, we're using the Midpoint Rule,
which means our rectangles would look like this.
So if we take a look at this midpoint formula for a moment,
notice how we have the sum of f of x sub i minus one
plus x sub i divided by two, this gives us the midpoint,
which we then sub into the function to find the height
of the rectangle, then we multiply this by delta x,
the width, to find the area of each rectangle.
Let's begin by determining delta x.
Delta x is equal to b minus a divided by n,
b is 3.5,
a is negative 0.5,
and n is given as four.
So notice that delta x is just four divided by four, or one.
Now that we have delta x, let's find
the endpoints of each interval.
So the first endpoint of the first interval
would be x sub zero, which will always
be just a, so negative 0.5.
And now to find x sub one, we just add delta x,
which is one, negative 0.5 plus one is positive 0.5,
x sub two would be .5 plus one, or 1.5,
x sub three, 1.5 plus one is 2.5, and finally
x sub four is 2.5 plus one, or 3.5.
We know we're done because notice how 3.5 is b.
So let's go ahead and set this up.
The integral from negative 0.5 to 3.5
of one-fourth x to the third dx
will be approximately equal to
the sum given by this midpoint formula.
So the first function value will be f of,
if i is one, notice how we'd have x sub zero
plus x sub one divided by two, or negative 0.5 plus 0.5
divided by two, that'll give us the height
of the first rectangle, times delta x, which is one,
plus f of, when i is two, we have x sub one plus x sub two
divided by two, to be 0.5 plus 1.5 divided by two,
there's the height of the second rectangle,
times delta x, which is one, plus f of, when i is three,
we have x sub two plus x sub three divided by two,
which would be 1.5 plus 2.5 divided by two,
this is the height of the third rectangle,
times delta x, which is one, plus f of,
when i is four, this will be the last value of i,
since n is four, we have x sub three
plus x sub four divided by two.
2.5 plus 3.5 divided by two times delta x, which is one.
Let's go ahead and simplify this.
Here we have f of, this would be zero times one,
plus, this would be f of, two divided by two is one,
so f of one times one, plus, this would be f of,
four divided by two is two, f of two times one,
and then finally plus f of, this would be six
divided by two, or f of three, times one.
So now we're going to find these function values
using the function f of x equals one-fourth x to the third,
and then find these products, and then find the sum.
But before we do this, let's look
at the rectangles this represents.
Here's a graph of our function.
Notice how the first interval
from negative .5 to 1.5 is here,
and we have .5 to 1.5 here,
1.5 to 2.5,
2.5 to 3.5.
Notice the midpoint of each interval
is determining the height of each rectangle.
Also notice, it looks like this first
rectangle has a height of zero.
Now let's go back and finish our calculation.
Well f of zero would be one-fourth
times zero cubed times one plus,
f of one would be one-fourth times
one cubed times one, plus, f of two
would be one-fourth times two cubed
times one, and f of three would be
one-fourth times three cubed times one.
And we'll finish this on the next slide.
This first product would be zero, then we have plus,
this would just be one-fourth, plus one-fourth
times two cubed, that's one-fourth times eight,
that's two times one is two, plus, here we have
one-fourth times three cubed, or one-fourth
times 27, which is 27-fourths.
It would just be one-fourth plus, two over one
is equivalent to eight-fourths, plus 27-fourths.
So we have 36-fourths, which is equal to nine.
So nine would be the approximation
for the def integral using the Midpoint Rule.
And since the function is not negative on this interval,
we could say we're using the area of these rectangles
to approximate the area under the function
on the interval from negative 0.5 to 3.5.
I hope you found this helpful.