We're asked to estimate
the area under the graph
of f of x equals the square root
of the quantity x plus 3
over the closed interval
from negative 3 to positive 3
using six approximating rectangles and midpoints.
So let's first mark off the closed interval
from negative 3 to positive 3,
which is here on the left and here on the right.
For the next step, let's determine the width
of each subinterval by determining delta x,
where delta x is equal to the quantity
b minus a divided by n.
In our case a is equal to negative 3,
b is equal to positive 3,
and because we are using six approximating rectangles,
n is equal to 6,
which gives us delta x
is equal to 3 minus negative 3 divided by 6,
which is equal to 6 divided by 6,
which is equal to 1.
Now we'll form six subintervals with the width of 1 unit
from negative 3 to positive 3.
Now that we have our subintervals,
we know x sub 0 is equal to negative 3,
x sub 1 is equal to negative 2,
x sub 2 is equal to negative 1, and so on,
where x sub 6 is equal to positive 3.
And now we'll sketch the six rectangles
that we'll use to approximate the area under the curve
bounded by the x-axis.
To do this, we'll determine the function values
at the midpoint of each subinterval.
The x-value in the middle of negative 3
and negative 2 is negative 2.5,
which means f of negative 2.5
is the height of the first rectangle.
So if I go up to the function
at x equals negative 2.5,
this would be the height of the first rectangle.
Let's also give the area of this rectangle
using function notation.
So M sub 6 is equal to,
the area of this first rectangle
is equal to the width times the height,
which is 1 times the function value
f of negative 2.5.
For the next subinterval,
the x-value in the middle of negative 2 and negative 1
is negative 1.5.
f of negative 1.5 is the height
of the second rectangle,
which would be here.
The area of the second rectangle is equal
to the width of 1 times the function value
f of negative 1.5 plus,
for the third subinterval,
the x-value in the middle of negative 1 and 0
is negative 0.5.
f of negative 0.5 is the height of the third rectangle,
which should be here in the middle of the subinterval.
The area is equal to the width of 1
times the function value f of negative 0.5.
The x-value in the middle of 0 and 1 is 0.5.
f of 0.5 is the height of the fourth rectangle,
which should be here
where the area is equal to 1 times f of 0.5.
For the fifth rectangle, the x-value in the middle
of 1 and 2 is 1.5.
f of 1.5 is the height of the fifth rectangle,
which is here in the middle of the subinterval.
The area of this rectangle is the width
of 1 times the function value f of 1.5.
And then for the sixth rectangle,
the x-value in the middle of 2 and 3 is 2.5.
f of 2.5 is the height of the sixth rectangle,
which should be here in the middle of the subinterval
where the area is equal to the width
of 1 times the function value f of 2.5.
Notice all the widths here are 1.
So if we want to match the formula here above,
we could say M sub 6 is equal to delta x,
which is 1 times the sum of the function values
f of negative 2.5
plus f of negative 1.5
plus f of negative 0.5
plus f of 0.5 plus f of 1.5
plus f of 2.5.
Let's go ahead and shade the area of the rectangles.
And now let's finish the calculation below.
We have M sub 6 is equal to 1
times the sum of the function values
where f of negative 2.5 is equal to the square root
of negative 2.5 plus 3,
which is the square root of 0.5,
plus f of negative 1.5 is equal to the square root
of negative 1.5 plus 3,
which is equal to the square to 1.5,
plus f of negative 0.5
is equal to the square root of negative 0.5 plus 3,
which is the square root of 2.5,
plus f of 0.5 is equal to the square root of 0.5 plus 3,
or the square to 3.5,
plus f of 1.5 is equal to the square root of 1.5 plus 3,
which is equal to the square root of 4.5.
And then finally,
f of 2.5 is equal to the square root of 2.5 plus 3,
which is the square root of 5.5.
Now we'll go to the calculator
and get our decimal approximation for this area,
and we'll round to four decimal places.
Of course, we can leave off this factor of 1
and just find the sum of the square roots.
To four decimal places, we have approximately 9.8503.
So using six approximating rectangles and midpoints,
we estimate the area under the curve
bounded by the x-axis over the closed interval
from negative 3 to positive 3
is approximately 9.8503 square units.
I hope you found this helpful.