In the previous section, the areas were easy to find because the regions involved simple geometric shapes. But suppose we wanted to find the area under a curved function like f(x)=x2 from x=a to x=b. Before we dive into the techniques to find the exact area under a curve, let’s start by approximating the area. One way is to divide the region into rectangles and then add the areas of those rectangles. This method is called a Riemann sum approximation.
Riemann sums to estimate area
There are 3 main types of Riemann approximations: the left, right, and midpoint Riemann sums, all of which use rectangles. The last way you’ll learn to estimate the area is by the Trapezoidal rule (using trapezoids). Although this method isn’t considered a Riemann approximation, the process is similar.
Here are the steps to finding a Riemann sum:
Divide the interval [a,b] into n subintervals, where each subinterval has equal width Δx:
Δx=nb−a
On each subinterval, choose a point xk to evaluate the function (which will vary depending on the type of Riemann sum being evaluated).
Multiply the function value f(xk) by the width Δx to get the area of a rectangle.
For a trapezoid, Δx is the height of each trapezoid while the function values are the bases.
Note: if the function value is negative, then the rectangle will be under the x-axis and the “area” will be negative. This is perfectly fine because what we’re looking to find is the net signed area (described in the previous section) which takes into account both positive and negative values.
Add the areas together.
Examples
Estimate the area under f(x)=x2 with n=4 subintervals on the interval [1,5] using a:
a) Right Riemann sum
b) Left Riemann sum
c) Midpoint Riemann sum
d) Trapezoidal sum
a) Right Riemann sum
Divide the interval [1,5] into n=4 subintervals, each with equal width:
Δx=nb−a
=45−1
=1
This step is the same for all of the sums.
Since n=4 and this is a right Riemann sum, draw 4 rectangles with the top-right corner of each touching the curve starting from the rightmost point of the interval, or b=5. Each has width Δx=1 and height corresponding to the function value.
Figure 6.2.1 Right Riemann sum
The right endpoints are x=5,4,3,2 and the function values are:
f(5)=25
f(4)=16
f(3)=9
f(2)=4
Then the right Riemann sum is
R4=Δx(f(5)+f(4)+f(3)+f(2))
=1(25+16+9+4)
=54 units2
b) Left Riemann sum
Just as in part a),
Δx=1
This time, each rectangle’s height is determined by where its top left corner touches the function, so we start from the leftmost point of the interval, or a=−2, and draw 4 rectangles to the right, each with the width Δx=1.
Figure 6.2.2 Left Riemann sum
The left endpoints are x=1,2,3,4 and the function values are:
f(1)=1
f(2)=4
f(3)=9
f(4)=16
Then the left Riemann sum is
L4=Δx(f(1)+f(2)+f(3)+f(4))
=1(1+4+9+16)
=30 units2
c) Midpoint Riemann sum
Just as in the previous parts,
Δx=1
The 4 rectangles are in these 4 subintervals, where the bottom corners touch the endpoint of each subinterval:
[1,2]
[2,3]
[3,4]
[4,5]
The height of each rectangle will be the function value of the midpoint of each interval. The midpoint is found by adding the ends and dividing by 2; for example,
Midpoint of [3,4]=
23+4=3.5
The midpoints of each subinterval in this example are x=1.5,2.5,3.5,4.5 and the function values are:
f(1.5)=2.25
f(2.5)=6.25
f(3.5)=12.25
f(4.5)=20.25
Figure 6.2.3 Midpoint Riemann sum
Then the midpoint Riemann sum is
M4=Δx(f(1.5)+f(2.5)+f(3.5)+f(4.5))
=1(2.25+6.25+12.25+20.25)
=41 units2
d) Trapezoidal sum
The area of a trapezoid is
A=2h(b1+b2)
The trapezoids are oriented sideways, where the height of each is Δx=1 and the lengths of the parallel bases are the function values.
Figure 6.2.4 Trapezoidal sum
For the leftmost trapezoid, the lengths of the parallel bases are f(1) and f(2).
For the trapezoid next to it, the parallel bases are f(2) and f(3).
Notice that when the areas are added, f(2) is added twice. For trapezoidal sums, all function values other than the endpoint ones will be added twice.
Then
T4=2Δx(f(1)+2f(2)+2f(3)+2f(4)+f(5))
=21(1+2(4)+2(9)+2(16)+25)
=42 units2
Unequal widths
In some AP exam problems, the width of each rectangle isn’t the same value of Δx. These problems will generally be in table form.
The rate of water, in gallons per minute, flowing into a tank can be modeled by function g(t), which is continuous and strictly increasing on the interval [0,10]. Select values are given below:
t
g(t)
0
3
2
5
5
10
7
12
10
15
a) Will a left or right Riemann sum underestimate the actual change in the amount of water?
b) Approximate the net change in the amount of water after 10 seconds using a left Riemann sum with the subintervals [0,2],[2,5],[5,7], and [7,10]. Include units.
c) Repeat part b) using a right Riemann sum with the same subintervals. Include units.
Solutions
a) Underestimation
(spoiler)
Because g(t) is always increasing, we can draw its approximate shape by connecting each point to the next. Doing this shows the left Riemann sum rectangles fall below the curve, meaning they give an underestimate of the actual area. On the other hand, right Riemann sums extend above the curve and overestimate.
b) Left Riemann sum
(spoiler)
The rectangles can be made as shown in the image below, with the units being that of g(t)×t, or gallons.
Figure 6.2.5 Left sum
The height of the 1st rectangle on the left is g(0)=3, the function value of the left endpoint. The width is the distance between the 2 partitions t=0 and t=2, or 2.
The 2nd rectangle has a height of g(2)=5, but the width this time is 3.
Continuing the same process with the remaining rectangles, the sum of the areas is:
A=2(3)+3(5)+2(10)+3(12)
=67 gallons
c) Right Riemann sum
(spoiler)
This time we start with the function value of the right endpoint, g(10), as the height of the 1st rectangle on the right.
Figure 6.2.6 Right sum
The sum of the areas of the rectangles is:
A=3(15)+2(12)+3(10)+2(5)
=109 gallons
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