Welcome to an example of numerical integration.
We want to use the Midpoint Rule with n equals three
to approximate the given definite integral.
Notice how we're integrating
on the interval from one to five
and since n is equal to three,
we'll be using three rectangles
to approximate this definite integral
and because we're using the Midpoint Rule,
we'll use the midpoint of each subinterval
to then determine the function value
which will be the height of each rectangle.
Let's begin by determining the width of each sub interval
which is equal to delta x which is b minus a divided by n,
where a and b are the limits of integration.
Delta x is equal to five minus one
divided by n which is three,
so the width of each subinterval
is going to be 4/3 of a unit.
Let's take a look at the graph of this.
Here's the graph of our function or the integrand
and because we're integrating from one to five,
one will be x sub zero.
And since delta x is 4/3,
we can add 4/3 to x sub zero
to find the end of the first subinterval.
One plus 4/3 would be somewhere in here.
This would be x sub one.
Add 4/3 again, we'd be somewhere in here
which would be x sub two.
Add 4/3 again and this would be x sub three.
Notice how this is also equal to b,
the upper limit of integration.
And now because we're using the Midpoint Rule,
let's go ahead and approximate the midpoint
of each subinterval.
Maybe the midpoint would be somewhere in here
for the first subinterval,
somewhere in here for the second subinterval
and somewhere in here for the third subinterval.
At these midpoints, we'll now determine the function value
which will be the height of each rectangle.
For this first subinterval,
this would be the height of the first rectangle
so the rectangle would look something like this.
For the next subinterval,
this would be the height of the rectangle.
Notice how the function value is negative
so the rectangle actually looks something like this.
Because the function value is negative there,
we can think of this as giving us a negative area
and then for this last subinterval,
the function value would be somewhere in here
so this would be our rectangle.
And again, because the function value is negative here,
we can think of this as being a negative area.
This area would give us a positive value,
these other two would give us negative values.
Now let's go ahead and calculate these three areas.
Again, delta x equals 4/3.
If we look at this formula here for the Midpoint Rule,
we have the sum of, in our case, i equals one to three
since n is three of f of x sub i minus one
plus x sub i divided by two.
This average here gives us the midpoint of each interval
which we then find the function value at
to determine the height of the rectangle,
then multiply this by delta x, the width of each rectangle.
We already know that x sub zero would be a or one,
x sub one since delta x equals 4/3 would be
one plus 4/3 or 7/3.
X sub two would be 7/3 plus 4/3 or 11/3
and x sub three would be 11/3 plus 4/3
which is 15/3 or just five.
Notice how this is b.
The integral from one to five
of negative x squared plus two x
is approximately equal to, using our formula here
we would have f of when i is one we have
x sub zero plus x sub one divided by two
which is one plus 7/3 divided by two.
This would be the height of the first rectangle
times delta x or 4/3
which is the width of the first rectangle.
Plus f of when i is two, we have x sub one plus x sub two
divided by two which is 7/3 plus 11/3
divided by two times delta x.
Again, this is the heart of the second rectangle
times the width plus f of when i is three,
we have x sub two plus x sub three divided by two
which is 11/3 plus five divided by two
times delta x which is 4/3.
Now we have f of, this will be one plus 7/3
which is 3/3 plus 7/3 which is 10/3.
And then divided by two is the same
as multiplying by 1/2.
10/3 times 1/2 simplifying here
would give us 5/3.
So, f of 5/3 times 4/3
plus here we have 7/3 plus 11/3,
that's 18/3 which is six.
Six divided by two is three,
so f of three times 4/3,
plus here we have 11/3 plus five.
11/3 plus 15/3 would be 26/3.
And then 26/3 divided by two or 26/3 times 1/2,
again this simplifies to 13/3.
We have f of 13/3 times 4/3.
Now let's go and find these function values,
to do this we'll use the calculator.
Let's go ahead and type in the function into y one
so I'll press y equals and then negative x squared,
plus two x.
And now I'm going to go back to the home screen,
second mode.
So to find f of 5/3, we can press vars right arrow,
enter, enter.
And then instead of f of 5/3,
we can type in y one of 5/3.
And this will give us our function value.
We press enter, we'll convert this back to a fraction
by pressing math enter, enter,
so we have 5/9 which means for this first product we'd have
5/9 times 4/3,
plus find f of three.
So again, vars, right arrow, enter, enter,
then in parenthesis three
which gives us negative three.
We have negative three which we'll add as
negative three over one times 4/3
plus f of 13/3.
So again, vars, right arrow, enter, enter
and then parenthesis 13/3.
Math enter, enter.
Negative 91/9.
Nothing simplifies here so we have 20/27 plus,
here we would just have negative four.
And here we would have negative 364/27.
This will come out to negative 452/27
or as a decimal approximation this is approximately
negative 16.7407.
I hope you found this explanation helpful.