Finding Areas with Riemann Sums

Warning: You must know summation notation, formulae, and operations do this lesson.

Area Under a Curve

In order to find the area between a given function's curve and the x-axis, or between two given curves, mathematical pioneers decided to divide the desired area into a finite number of rectangles with equal or unequal bases and then sum the areas of these rectangles.

This method yields a close estimate of the desired area however, it also includes or excludes spaces, since the rectangles don't exactly fit the area defined by the curve. There are always little bits more or less than the area we want.

They solved this problem by limiting n -- the number of rectangles -- to infinity. This makes each rectangle's base so teeny that the included or excluded triangles are eliminated and we get the exact area under or between the curve(s).

So we'll deal with two types of questions on finding area from fundamental principles. In the first type, we'll divide the area up into rectangles with equal bases and in the second type, we'll make a rectangular partition of the area with predefined borders. In the first case, we'll get the exact area we want because we will limit n -- the number of rectangles -- to infinity. In the second case, we will find an estimate for the area since we will not limit anything to infinity. The difference should be obvious from the wording of the question. In the first case, we'll be asked to evaluate the area, in the second, the question will define the border values of x and will ask us to estimate the area under the curve.

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Riemann Sums

Say we want to find the area under the curve f (x) = x² from x = 0 to x = 5.
We divide the 5 units of the x-axis into n rectangles with equal bases = 5/n.
Now we need to find the x-value we'll use to determine the height of the rectangles, so we need an expression for the x value on the axis at the right end of each rectangle's base.

We see that the right endpoints of each rectangle's base is:

in rectangle #1,

in rectangle #2,

so in the " i" th rectangle the x value is with i between 1 and n.
where i is the order number of each rectangle. (1st, 2nd, ... nth)
Since , , the height

The area of a rectangle is height times base, so the area of the i th rectangle here is:

If we sum the above expression over the interval from 0 to 5, we will have the area.

The problem however, is that we find not only the area under the curve, but also the area of the triangles above the curve. To eliminate these, we limit n, the number of rectangles, to infinity. This makes the bases teeny since there are so many of them and so gives us the area between the
curve f (x) = x2 and the x-axis. Note that limiting n to infinity makes approach zero.

So our final expression for the area we're seeking is:

We have a formula for the summation of i 2 , we apply it now and then take the limit as n approaches infinity. Recall, to do this we divide our fraction through by the highest power of the variable, then set the variable equal to infinity.

When we switch the denominators and find the limit for the 2nd fraction (with all the n's in it),
we get 2, since the highest power of n is n3 and the coefficient of the numerator's n3 term is 2.

When we reduce the fraction, the answer is .

Now say we wish to find the area between the x-axis and the curve f (x) = x2
between x = 2 and x = 5.
The process is almost identical except now, , and ,
so

When we multiply by f (x) we get:

So our area this time is:

When we substitute the formulae for the summations of i and , we get 12 + 18 + 9 = 39.

Now let's do it by integration. We evaluate

Note that the area is a summation of the product f (x i ) times which is f (x) dx. Now you know why there's dx in every integrand. It represents the base of the rectangle.

Notation

When dealing with partitions, there are a number of notation systems used to denote the
values of x that define the borders and heights for the rectangles.

Generally, the border values for the partitions are denoted x i , x j or x k .

Some texts use x* for the x-value used to find the height or f (x*), others use wk for this.

Riemann Sums with Partitions

Now let's evaluate a partition of the area under the curve f (x) = x² + 1 on the interval from
a = 0 to b = 6 and with n = 5 and x 1 = 1, x 2 = 2, x 3 = 3, x 4 = 4, x 5 = 6.
We'll use the midpoint of the partition as x * or wk .

As we see from the x-values, the length of the 1st four bases = 1, the 5th = 2.
The midpoint values are:

 w1 = 0.5 w2 = 1.5 w3 = 2.5 w4 = 3.5 w5 = 5

We need values for f(x i ) since those are the heights of the rectangles.

 f (w1) = 1.25 f (w2) = 3.25 f (w3 ) = 7.25 f (w4 ) = 13.25 f (w5) = 26

The sum = 1.25 + 3.25 + 7.25 + 13.25 + 2(26) = 77 (units) ² , so this our estimate for the area.

Though it is not obvious in the diagram, we're also finding the area of the little triangles above the curve y = x 2 + 1. To eliminate this extra area, Riemann stated that we need only limit the largest base to zero in order to limit all bases to zero.

In our case however, we need only estimate the Riemann sum described in the question.

There are times when we must use the left or right endpoint of the base interval. The question will say which one to use. If there are no specific instructions, we should always use the right endpoint since it's easiest to find.

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Should the curve in question cross the x-axis at some point on the given interval, we must take into account that the heights of the rectangles below the axis will all be negative. Since area can't be negative, we break it up into two parts. We multiply all negative heights by – 1 to make them positive, we find the two areas separately and then sum them. For this reason, it's always best to make a diagram of the situation in question and to check the interval for intercepts.

Here, we'll find the area under f (x) from x =0 to x = 5 and we'll add it to the area under
g(x) from x = 5 to x = 7, since multiplying the function g(x) by – 1 flips the curve
to generate the identical area above the x-axis.

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Summary

To find the exact area under a curve f (x):

1. Divide the x-axis interval (a, b) into n rectangles with equal bases.

2. Find an expression for , the length of each base, .

3. Find an expression for x i , (with i from 1 to n) , the right endpoint of the i th rectangle.

4. Find an expression for f ( x i ) , the height of the i th rectangle.

5. Multiply height f ( x i ) by base , to get an expression for the area of i th rectangle.

6. Sum the formula in step 5 from i = 1 to n. Substitute the summation formulae.

7. Limit n to infinity to eliminate the under- or over-hang area.

To estimate area under a curve using a Partition:

1. Draw a diagram using the border values defined in the question.

2. Make a table of values for , x i , and f ( x i ) at the specified end or mid points.

3. Multiply base by height for each rectangle then sum the areas.

1) Use a Riemann sum to find the area between f (x) and the x-axis, over the given interval.

 a) f (x) = 5 – x²with a = 0, and b = 2 b) f (x) = x³ + 8with a = 0, and b = 5 c) f (x) = x² + 1with a = 1, and b = 3

| (summation formulae) | (solutions)

2) Estimate the area under the curve f (x) = x² + 1 from a = 0 to b = 6 with the
regular partition in which n = 6 and x 1 = 1, x 2 = 2, x 3 = 3, x 4 = 4, x 5 = 5, and x 6 = 6.
Use the midpoint of the partitions as x * or wk . (make a diagram!)

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1) a)

We see that , so which makes the height

The area of the i th rectangle is

The total area then is

Now, as before, we switch the denominators and evaluate the summation of i².

So we get square units.

When we integrate the function from 0 to 2 we get exactly the same value for the area.

b)

We see that , so which makes the height

The area of the i th rectangle is

The total area then is

Now, as before, we switch the denominators and evaluate the summation of i³.

So we get square units.

When we integrate the function from 0 to 5 we get exactly the same value for the area.

c)

We see that , but since we start at 1,
which makes the height f ( x i ) =

The area of the i th rectangle is

The area then is

Now, as before, we switch the denominators and evaluate the summation of i and i².

So we get square units.

When we integrate the function from 1 to 3 we get exactly the same value for the area.

2) use diagram above

The length of each base = 1
The midpoint values are:

 w1 = 0.5 w2 = 1.5 w3 = 2.5 w4 = 3.5 w5 = 4.5 w6 = 5.5

We need values for f (x i ) since those are the heights of the rectangles.

 f (w1) = 1.25 f (w2) = 3.25 f (w3 ) = 7.25 f (w4 ) = 13.25 f (w5) = 21.25 f (w6) = 30.25

The total = 1.25 + 3.25 + 7.25 + 13.25 + 21.25 + 30.25 = 76.5 units ² , so this our estimate for the area.

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 a) b) c) d) e) f) g)

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