Introduction to Riemann Sums
I’m convinced the reason they teach you Riemann Sums and other methods to find areas is to have you “appreciate” what our former mathematicians had to go through before things got easier. I’m the first to admit that I’m not a fan of working with them, just because they are so tedious.
These methods can be used to approximate the area between a curve and the $ \boldsymbol {x}$-axis, which can be acquired much easier by just taking the integral of the function between two different $ x$-values (we’ll do this in the Definite Integration section). But, alas, we have to learn these more difficult methods first.
Important note: “Area under a curve” assumes the curve is above the $ x$-axis. For curves that are both under and above the $ x$-axis, the areas must be computed separately and added together. A Definite Integral computes the area above the $ x$-axis minus any area below the $ x$-axis.
Let’s say we wanted to get the area of the region between $ x=1$ and $ x=2$ for certain “curved” functions above the $ x$-axis. To get an approximation, we could just add up little rectangles that we can form between $ x=1$ and $ x=2$ that are under the curve and above the $ x$-axis. And do you see how the more rectangles we use (the “$ n$”), the more accurate the total area will be?
Do you also see how, depending on whether the upper left or upper right (or midpoint) of the rectangles touch the curve, we’ll get slightly different areas? For example, for some functions, the “right-hand” sums (where the right-hand sides of the rectangles touch the function) over-approximates the area (upper sum, meaning the rectangles go above the actual curve), and for others, it under-approximates the area (lower sum, meaning the rectangles go below the actual curve). Note that “$ n$” represents the number of rectangles.
How does all this relate to Calculus? Soon we’ll learn to use Integration to get the area between a curve and the $ x$-axis, but again, to “appreciate the math”, we’ll first learn how to approximate the area using upper, lower, and midpoint sums.
Using Upper and Lower Sums to Approximate Area
Here’s an example of using a lower sum to estimate area; in this case, it’s a left-hand sum, since the upper left part of the rectangle touches the curve. The notation can be scary looking, but it’s not that bad.
Using Lower Sum to Approximate Area between x = 1 and x = 2:
Function $ y={{x}^{2}}$, $ n=8$ intervals, left-hand sum: 
Note that for interval $ [a,b]$, to get the width ($ \Delta x$) of each rectangle, subtract the small number $ (a)$ from the larger number $ (b)$ and then divide by the number of intervals $ (n)$. Thus, each width ($ \Delta x$) is $ \displaystyle \frac{{b-a}}{n}=\frac{{2-1}}{8}=\frac{1}{8}$.
To get each height, use the function $ y={{x}^{2}}$, since the $ \boldsymbol {y}$-value gives us each height. For the lower sum, we have a left-hand sum for this function, and we need the $ \boldsymbol {y}$ that starts with $ x=1$ and ends before $ x=2$; in fact, it ends at $ \displaystyle x=1\frac{7}{8}$. (Note that if were getting a right-hand sum, we’d start with $ \displaystyle x=1\frac{1}{8}$ and end with $ x=2$). You can see that the left-hand estimation will be an underestimate.
To get our $ y$-values, plug in the $ x$-values $ \displaystyle 1,\,\,1\frac{1}{8},\,\,1\frac{2}{8},\,…\,\,1\frac{7}{8}$ into $ y={{x}^{2}}$, and then multiply each $ y$-value by $ \displaystyle \frac{1}{8}$ to get the area of each of the 8 rectangles. Add these individual areas up to get the total area:
$ \displaystyle \begin{align}\text{Estimated Area}&=\frac{1}{8}\cdot f\left( 1 \right)+\frac{1}{8}\cdot f\left( {1\frac{1}{8}} \right)+\frac{1}{8}\cdot f\left( {1\frac{2}{8}} \right)+\frac{1}{8}\cdot f\left( {1\frac{3}{8}} \right)+\frac{1}{8}\cdot f\left( {1\frac{4}{8}} \right)\\&\,\,\,\,\,\,\,+\frac{1}{8}\cdot f\left( {1\frac{5}{8}} \right)+\frac{1}{8}\cdot f\left( {1\frac{6}{8}} \right)+\frac{1}{8}\cdot f\left( {1\frac{7}{8}} \right)\\&=\frac{1}{8}\cdot {{1}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{1}{8}} \right)}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{2}{8}} \right)}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{3}{8}} \right)}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{4}{8}} \right)}^{2}}\\&\,\,\,\,\,\,\,+\frac{1}{8}\cdot {{\left( {1\frac{5}{8}} \right)}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{6}{8}} \right)}^{2}}+\frac{1}{8}\cdot {{\left( {1\frac{7}{8}} \right)}^{2}}\\&=\frac{1}{8}\left[ {{{1}^{2}}+{{{\left( {\frac{9}{8}} \right)}}^{2}}+{{{\left( {\frac{{10}}{8}} \right)}}^{2}}+{{{\left( {\frac{{11}}{8}} \right)}}^{2}}+{{{\left( {\frac{{12}}{8}} \right)}}^{2}}+{{{\left( {\frac{{13}}{8}} \right)}}^{2}}+{{{\left( {\frac{{14}}{8}} \right)}}^{2}}+{{{\left( {\frac{{15}}{8}} \right)}}^{2}}} \right]\\&=\left( {\frac{1}{8}} \right)\left( {\frac{{276}}{{15}}} \right)\approx 2.1484\end{align}$
You can see that it’s an underestimate, since the actual value is $ \displaystyle 2\frac{1}{3}\approx 2.33$.
Using Midpoint Rule to Approximate Area
The midpoint rule uses sums that touch the function at the center of the rectangles that are under the curve and above the $ x$-axis. We compute the area approximation the same way, but evaluate the function right in between (the midpoint of) each of the rectangles; this will be height of the rectangles (the “$ y$”). We’ll compute the $ \Delta x$ the same way. We’ll see an example below.
Upper, Lower, and Midpoint Sums Problems
Here are examples of upper, lower, and midpoint sum problems:
| Lower, Upper and Midpoint Sum Problems | Solution | ||||||||||||
| Use lower and upper sums to estimate the area of the region bounded by $ y=\sqrt{x}+2$ and the $ x$-axis between $ x=0$ and $ x=1$, using $ 4$ intervals.
| The width ($ \Delta x$) of each rectangle is $ \displaystyle \frac{{b-a}}{n}=\frac{{1-0}}{4}=\frac{1}{4}$.
For lower sums, use the left-hand rule and use $ x$-values $ \displaystyle 0,\,\frac{1}{4},\,\frac{2}{4},$ and $ \displaystyle \frac{3}{4}$ in the function $ y=\sqrt{x}+2$ to get the heights of the rectangles. Again, the width of the rectangles is $ \displaystyle \frac{1}{4}$. Sum up the areas of these four rectangles (underestimate): $ \displaystyle \begin{align}\text{area}&\approx \frac{1}{4}\cdot f\left( 0 \right)+\frac{1}{4}\cdot f\left( {\frac{1}{4}} \right)+\frac{1}{4}\cdot f\left( {\frac{2}{4}} \right)+\frac{1}{4}\cdot f\left( {\frac{3}{4}} \right)\\&=\frac{1}{4}\left[ {\left( {\sqrt{0}+2} \right)+\left( {\sqrt{{\frac{1}{4}}}+2} \right)+\left( {\sqrt{{\frac{2}{4}}}+2} \right)+\left( {\sqrt{{\frac{3}{4}}}+2} \right)} \right]\\&\approx \left( {\frac{1}{4}} \right)10.073\approx 2.518\end{align}$
For upper sums, use the right-hand rule and use $ x$-values $ \displaystyle \frac{1}{4},\,\,\frac{2}{4},\,\text{ }\frac{3}{4}\text{,}$ and $ 1$ in the function $ y=\sqrt{x}+2$ to get the heights of the rectangles. Again, the width of the rectangles is $ \displaystyle \frac{1}{4}$. Sum up the areas of these four rectangles (overestimate): $ \displaystyle \begin{align}\text{area}&\approx \frac{1}{4}\cdot f\left( {\frac{1}{4}} \right)+\frac{1}{4}\cdot f\left( {\frac{2}{4}} \right)+\frac{1}{4}\cdot f\left( {\frac{3}{4}} \right)+\frac{1}{4}\cdot f\left( 1 \right)\\&=\frac{1}{4}\left[ {\left( {\sqrt{{\frac{1}{4}}}+2} \right)+\left( {\sqrt{{\frac{2}{4}}}+2} \right)+\left( {\sqrt{{\frac{3}{4}}}+2} \right)+\left( {\sqrt{1}+2} \right)} \right]\\&\approx \left( {\frac{1}{4}} \right)11.073\approx 2.768\end{align}$ | ||||||||||||
| The function $ f$ is continuous on the closed interval $ \left[ {2,13} \right]$ and has values show in the table. Using the intervals $ \left[ {2,3} \right]$, $ \left[ {3,5} \right]$, $ \left[ {5,9} \right]$, and $ \left[ {9,13} \right]$, what is an approximation of using a left Riemann sum (without graphing).
| This is a common AP Calculus problem that requires some thinking! Since we’re obtaining a left sum, use the first 4 points (if it were a right sum, it’d be the last 4 points).
Note that the difference between the $ x$’s are $ \displaystyle 1\text{ }\left( {3-2} \right),\text{ }2\text{ }\left( {5-3} \right),\text{ }4\text{ }\left( {9-5} \right),\text{ and }4\text{ }\left( {13-9} \right)$. These are the widths of the rectangles. The lengths of the rectangles are the $ f\left( x \right)$ values.
Thus, $ \displaystyle \int\limits_{{13}}^{2}{{f\left( x \right)}}\,dx\approx 1\left( 6 \right)+2\left( {-2} \right)+4\left( {-1} \right)+4\left( 3 \right)=10$. | ||||||||||||
| Use the midpoint rule to estimate the area of the region bounded by $ y={{x}^{2}}+2x$ and the $ x$-axis between $ x=1$ and $ x=5$, using $ 4$ intervals.
Note: Actual area $ \approx 65.333$. | The width ($ \Delta x$) of each rectangle is $ \displaystyle \frac{{b-a}}{n}=\frac{{4-0}}{4}=1$.
For the midpoint rule, use midpoint of each rectangle and put that $ x$-value in the function. Evaluate the function $ y={{x}^{2}}+2x$ where $ x$ starts at $ \boldsymbol {a}$ ($ 1$) and goes to $ \boldsymbol {b}$ ($ 5$), divided in $ 4$ intervals. These values are: $ \displaystyle \frac{3}{2},\,\frac{5}{2},\,\frac{7}{2}\,\text{,}$ and $ \displaystyle \frac{9}{2}$. Again, the width of the rectangles is $ 1$. Sum up the areas of these four rectangles:
$ \displaystyle \begin{align}\text{area}&\approx 1\cdot f\left( {\frac{3}{2}} \right)+1\cdot f\left( {\frac{5}{2}} \right)+1\cdot f\left( {\frac{7}{2}} \right)+1\cdot f\left( {\frac{9}{2}} \right)\\&=\left[ {{{{\left( {\frac{3}{2}} \right)}}^{2}}+2\left( {\frac{3}{2}} \right)} \right]+\left[ {{{{\left( {\frac{5}{2}} \right)}}^{2}}+2\left( {\frac{5}{2}} \right)} \right]+\left[ {{{{\left( {\frac{7}{2}} \right)}}^{2}}+2\left( {\frac{7}{2}} \right)} \right]+\left[ {{{{\left( {\frac{9}{2}} \right)}}^{2}}+2\left( {\frac{9}{2}} \right)} \right]\\&=65\end{align}$ |
Note: Actual area $ \approx 2.667$.
Trapezoidal Rule
The Trapezoidal Rule, another variation of a Riemann Sum, evaluates areas under a curve (above the $ x$-axis), but uses trapezoids instead of rectangles. A trapezoid is a quadrilateral (four-sided figure) with two sides parallel (the bases) and two sides not parallel. Remember from Geometry how to get the area of a trapezoid; here’s how we’ll use it in Calculus:
| Trapezoidal Rule The area of a trapezoid is $ \displaystyle A=\frac{{{{b}_{1}}+{{b}_{2}}}}{2}\cdot h$.
Think of multiplying the height by the “average” of the two base values, which are parallel.
Here’s what using the Trapezoidal Rule to estimate area under a curve (above the $ x$-axis) looks like; notice how the trapezoids are turned on their sides, with each of the $ x$-intervals the same. |  |
We’ll want to create little trapezoids (with bases as $ y$-values) under the curve and above the $ x$-axis between two $ x$ points, and then add up all those areas. Here is the formal definition of the Trapezoidal Rule, as it relates to Definite Integration, which we’ll learn about in the next section:
The Trapezoidal Rule:
Let $ f$ be continuous on interval $ \left[ {a,\,b} \right]$. We can approximate $ \displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx$ by the Trapezoidal Rule:
$ \displaystyle \begin{align}\int\limits_{a}^{b}{{f\left( x \right)}}\,dx\,\,\approx \,\,\frac{{b-a}}{{2n}}\left[ {f\left( {{{x}_{0}}} \right)+2f\left( {{{x}_{1}}} \right)+…+2f\left( {{{x}_{{n-1}}}} \right)+f\left( {{{x}_{n}}} \right)} \right],\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{where }f\left( {{{x}_{0}}} \right)=f\left( a \right),\,\,\,f\left( {{{x}_{1}}} \right)=f\left( {a+\frac{{b-a}}{n}} \right),\,\,\text{and so on}\end{align}$
This is because each trapezoid is “on their side” and their height is $ \displaystyle \frac{{b-a}}{n}$ and bases are consecutive $ \displaystyle f\left( x \right)$ values. When we add up all the “averages” of the bases $ \displaystyle \frac{{f\left( {x{}_{1}} \right)+f\left( {x{}_{2}} \right)}}{2},\frac{{f\left( {x{}_{2}} \right)+f\left( {{{x}_{3}}} \right)}}{2}$ and so on, we end up with dividing by $ 2$ (thus the $ \displaystyle \frac{{b-a}}{{2n}}$ on the outside) and having twice all the $ \displaystyle f\left( x \right)$ values, except for the first and last.
Here are some Trapezoidal Rule problems. Note that when we have “$ n$” intervals, we will get “$ n+1$” $ f(x)$ values.
| Trapezoidal Rule Problem and Graph | Solution |
| Using the trapezoidal rule with 4 equal subintervals, approximate the integral: $ \displaystyle \int\limits_{0}^{1}{{\sqrt{{2+{{x}^{3}}}}}}\,dx$.
| Divide $ 0$ to $ 1$ into $ 4$ intervals, so $ f\left( x \right)$’s are $ \displaystyle 0,.25,.5,.75,\text{and }1$: $ \begin{align}\int\limits_{a}^{b}&{{f\left( x \right)}}\,dx\approx \frac{{b-a}}{{2n}}\left[ {f\left( {{{x}_{0}}} \right)+2f\left( {{{x}_{1}}} \right)+…+2f\left( {{{x}_{{n-1}}}} \right)+f\left( {{{x}_{n}}} \right)} \right]\\&=\frac{{1-0}}{{2\cdot 4}}\left[ {f\left( 0 \right)+2f\left( {.25} \right)+2f\left( {.5} \right)+f\left( {.75} \right)+f\left( 1 \right)} \right]\\&=\frac{1}{8}\left( {\sqrt{{2+{{{\left( 0 \right)}}^{3}}}}+2\sqrt{{2+{{{\left( {.25} \right)}}^{3}}}}+2\sqrt{{2+{{{\left( {.5} \right)}}^{3}}}}+2\sqrt{{2+{{{\left( {.75} \right)}}^{3}}}}+\sqrt{{2+{{{\left( 1 \right)}}^{3}}}}} \right)\\&\approx \frac{1}{8}\left( {12.014} \right)\approx 1.502\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(Actual value}\approx \text{ 1}\text{.497)}\end{align}$ |
| Using the trapezoidal rule with $ 3$ equal subintervals, approximate the integral: $ \displaystyle \int\limits_{{\frac{\pi }{2}}}^{\pi }{{\sqrt{{2x}}}}\sin \left( x \right)\,dx$.
| Divide $ \displaystyle \frac{\pi }{2}$ to $ \pi $ into $ 3$ intervals, so $ \displaystyle \frac{\pi }{2},\,\,\frac{{4\pi }}{6}\,\left( {\frac{{2\pi }}{3}} \right),\,\frac{{5\pi }}{6},$ and $ \pi $. (To get these, subtract $ \displaystyle \frac{\pi }{2}$ from $ \pi $ to get $ \displaystyle \frac{\pi }{2}$, and then divide by $ 3$ to get $ \displaystyle \frac{\pi }{6}$. The starting point is $ \displaystyle \frac{\pi }{2}$, and add $ \displaystyle \frac{\pi }{6}$ up to $ \pi $.) $ \displaystyle \begin{align}\int\limits_{a}^{b}&{{f\left( x \right)}}\,dx\approx \frac{{b-a}}{{2n}}\left[ {f\left( {{{x}_{0}}} \right)+2f\left( {{{x}_{1}}} \right)+…+2f\left( {{{x}_{{n-1}}}} \right)+f\left( {{{x}_{n}}} \right)} \right]\\&=\frac{{\pi -\frac{\pi }{2}}}{{2\cdot 3}}\left[ {f\left( {\frac{\pi }{2}} \right)+2f\left( {\frac{{2\pi }}{3}} \right)+2f\left( {\frac{{5\pi }}{6}} \right)+f\left( \pi \right)} \right]\\&=\frac{\pi }{{12}}\left[ \begin{array}{l}\left( {\sqrt{{2\left( {\frac{\pi }{2}} \right)}}\sin \left( {\frac{\pi }{2}} \right)} \right)+2\left( {\sqrt{{2\left( {\frac{{2\pi }}{3}} \right)}}\sin \left( {\frac{{2\pi }}{3}} \right)} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2\left( {\sqrt{{2\left( {\frac{{5\pi }}{6}} \right)}}\sin \left( {\frac{{5\pi }}{6}} \right)} \right)+\left( {\sqrt{{2\left( \pi \right)}}\sin \left( \pi \right)} \right)\end{array} \right]\\&\approx \frac{\pi }{{12}}\left( {7.60559} \right)\approx 1.99114\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(Actual value }\approx \text{ 2}\text{.0617)}\end{align}$ |
You may be given the actual function, or you may be given values at certain points so you’d have to “build” the trapezoids. Let’s do a problem where we don’t have equal intervals:
Trapezoidal Rule Problem
A park ranger needs to know the volume of a pond that she’s stocking with fish. The average depth of the pond is 25 feet, and the width of the pond at 100 feet intervals is given in the table below. Use the trapezoidal rule with 5 intervals to approximate the volume of this pond.
| Distance from west end of the pond | 0 | 100 | 175 | 275 | 390 | 500 |
| Distance across pond | 0 | 150 | 130 | 140 | 160 | 0 |
Solution: First use the Area of a Trapezoid formula to add up the area of all the trapezoids. For each trapezoid: $ \displaystyle A=\frac{{{{b}_{1}}+{{b}_{2}}}}{2}\cdot h$. The bases are the parallel lines above, and the height is the distance between the bases.
$ \begin{align}\text{Total}\,\text{Area}&=\frac{{0+150}}{2}\cdot \left( {100-0} \right)+\frac{{150+130}}{2}\cdot \left( {175-100} \right)+\frac{{130+140}}{2}\cdot \left( {275-175} \right)\\\,&\,\,\,\,\,\,+\frac{{140+160}}{2}\cdot \left( {390-275} \right)+\frac{{160+0}}{2}\cdot \left( {500-390} \right)\\\,&=\frac{1}{2}\left( {150\cdot 100+280\cdot 75+270\cdot 100+300\cdot 115+160\cdot 110} \right)=\frac{1}{2}\cdot 115100=57550\end{align}$
Multiply this by the depth of the pond ($ 25$ feet) to get Volume $ = 1,438,750$ feet3.
Area by Limit Definition
The area by limit definition takes the same principals we’ve been using to find the sums of rectangles to find area, but goes one step further. We’ll be finding the area between a function and the $ x$-axis between two $ x$ points, but doing it in a way that we’ll use as many rectangles as we can by taking the limit of the number of rectangles as that limit goes to $ \infty $. This way we’ll get the most accurate area we can, since the more rectangles we have, the closer we get the true area.
Again, this is a tough concept to grasp, but we’ll just use a formula that (hate to say it!) you’ll want to memorize how to use, as shown below. Note that $ \Sigma $ (the sigma sign) means to start with the first value, plug it in and go up to the last value, and taking the sum of all of those terms. Notice from the picture that this formula is closest to the midpoint rule.
Here’s the area by limit formula:
| Definition of the Area of a Region by a Limit
Let $ f$ be continuous and above the $ x$-axis (non-negative) on interval $ \left[ {a,\,b} \right]$. The area of the region bounded by $ f$, the $ x$-axis, and vertical lines at $ x=a$ and $ x=b$ is: $ \displaystyle \text{Area}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f\left( {{{c}_{i}}} \right)\cdot \Delta x}}$, where $ \displaystyle \Delta x=\frac{{b-a}}{n}$, and $ {{x}_{{i-1}}}\le {{c}_{i}}\le {{x}_{i}}$. Note that we will learn later that this area is $ \displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx$. |  |
Here are the formulas that will be used:
Area between $ x=a$ and $ x=b$ for function $ f$ is $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\,\sum\limits_{{i=1}}^{n}{{f\left[ {\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)\cdot \left( {\frac{{b-a}}{n}} \right)} \right]}}$ (“memorize” this one!)
Height x Width
To simplify and get rid of summation signs, use these summation formulas (usually given, but if not, memorize):
$ \displaystyle \sum\limits_{{i=1}}^{n}{{c=cn\,\,(c\text{ is a constant)}\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{i=\frac{{n\left( {n+1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{2}}=\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{3}}=\frac{{{{n}^{2}}{{{\left( {n+1} \right)}}^{2}}}}{4}\,\,}}$
All this looks a little scary, and it’s a bit tedius, but here the steps:
- Use the area formula $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\,\sum\limits_{{i=1}}^{n}{{f\left[ {\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)\cdot \left( {\frac{{b-a}}{n}} \right)} \right]}}$ with the given function and interval, and then simplify as much as you can.
- Separate the summations, if needed, and then leave only the “$ i$’s” in the summation (by moving everything else to the outside).
- Use the above summation formulas to turn “$ i$’s” into “$ n$’s”.
- Simplify and take the limit of what’s left (remember that $ \displaystyle \underset{{n\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{n}^{r}}}}=0$).
Also, don’t forget these useful summation properties: $ \displaystyle \sum\limits_{{i=1}}^{n}{{\left( {{{a}_{i}}+{{b}_{i}}} \right)\,=}}\sum\limits_{{i=1}}^{n}{{{{a}_{i}}\,}}+\sum\limits_{{i=1}}^{n}{{{{b}_{i}}\,}}\,\,\,\,\,\,\,\,\,\,\,\sum\limits_{{i=1}}^{n}{{c{{a}_{i}}\,=}}c\sum\limits_{{i=1}}^{n}{{{{a}_{i}}\,}}$.
Here’s an example:
Area by Limit Definition Problem:
Use the limit process with $ n$ rectangles to find the area of the region between $ f\left( x \right)={{x}^{2}}+4$ and the $ x$-axis over the interval $ [0,2]$.
Solution:
First determine the height and width of each rectangle, with $ f\left( x \right)={{x}^{2}}+4$ ($ a=0$ and $ b=2$).
Height = $ \displaystyle f\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)=f\left( {0+\frac{{\left( {2-0} \right)i}}{n}} \right)=f\left( {\frac{{2i}}{n}} \right)={{\left( {\frac{{2i}}{n}} \right)}^{2}}+4=\frac{{4{{i}^{2}}}}{{{{n}^{2}}}}+4$; Width = $ \displaystyle \Delta x=\frac{{b-a}}{n}=\frac{{2-0}}{n}=\frac{2}{n}$.
Use the area formula: $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\text{height}\,\cdot \,\text{width}} \right)=}}\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left[ {\left( {\frac{{4{{i}^{2}}}}{{{{n}^{2}}}}+4} \right)\left( {\frac{2}{n}} \right)} \right]}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\frac{{8{{i}^{2}}}}{{{{n}^{3}}}}+\frac{8}{n}} \right)}}$
Now that we’ve simplified, separate the summations if needed, and bring out the numbers and $ n$’s to the outside of the summations (for the second term, leave the summation with a “1”, which is a constant):
$ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\frac{{8{{i}^{2}}}}{{{{n}^{3}}}}+\frac{8}{n}} \right)}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\sum\limits_{{i=1}}^{n}{{\frac{{8{{i}^{2}}}}{{{{n}^{3}}}}+}}\sum\limits_{{i=1}}^{n}{{\frac{8}{n}}}} \right)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{8}{{{{n}^{3}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{2}}}}+\frac{8}{n}\,\sum\limits_{{i=1}}^{n}{1}} \right)$
Remember the summation formulas:
$ \displaystyle \sum\limits_{{i=1}}^{n}{{c=cn\,\,(c\text{ is a constant)}\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{i=\frac{{n\left( {n+1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{2}}=\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{3}}=\frac{{{{n}^{2}}{{{\left( {n+1} \right)}}^{2}}}}{4}\,\,}}$
Get rid of the $ \Sigma $’s by turning all the $ \Sigma i$’s into $ n$’s using the formulas above. Then simplify, and take the limits, remembering that $ \displaystyle \underset{{n\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{n}^{r}}}}=0$.
$ \require {cancel} \displaystyle \begin{align}\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{8}{{{{n}^{3}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{2}}}}+\frac{8}{n}\,\sum\limits_{{i=1}}^{n}{1}} \right)&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left[ {\frac{8}{{{{n}^{3}}}}\left( {\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}} \right)+\left[ {\frac{8}{n}\left( {1n} \right)} \right]} \right]\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left[ {\frac{{{}^{4}\cancel{8}}}{{{}_{{{{n}^{2}}}}\cancel{{{{n}^{3}}}}}}\left( {\frac{{\cancel{n}\left( {2{{n}^{2}}+3n+1} \right)}}{{{{{\cancel{6}}}_{3}}}}} \right)+\left( {\frac{{8\cancel{n}}}{{\cancel{n}}}} \right)} \right]\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{8{{n}^{2}}+12n+4}}{{3{{n}^{2}}}}+8} \right)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{8}{3}+{{{\cancel{{\frac{4}{n}}}}}^{0}}+{{{\cancel{{\frac{4}{{3{{n}^{2}}}}}}}}^{0}}+8} \right)=\frac{8}{3}+8=10\frac{2}{3}\end{align}$
Here’s another Area by Limit Definition problem. This can get a little messy, so you have to be careful. And I promise you that we’ll have a much easier way to get these areas soon! They just want you to “appreciate the math” at this point:
Area by Limit Definition Problem
Use the limit process with $ n$ rectangles to find the area of the region between $ f\left( x \right)={{x}^{3}}-1$ and the $ x$-axis over the interval $ [2,5]$.
Solution:
First determine the height and width of each rectangle, with $ f\left( x \right)={{x}^{3}}-1$ ($ a=2$, and $ b=5$).
Height = $ \displaystyle \begin{align}f\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)&=f\left( {2+\frac{{\left( {5-2} \right)i}}{n}} \right)=f\left( {2+\frac{{3i}}{n}} \right)={{\left( {2+\frac{{3i}}{n}} \right)}^{3}}-1={{\left( {2+\frac{{3i}}{n}} \right)}^{2}}\left( {2+\frac{{3i}}{n}} \right)-1\\&=\left( {4+\frac{{12i}}{n}+\frac{{9{{i}^{2}}}}{{{{n}^{2}}}}} \right)\left( {2+\frac{{3i}}{n}} \right)-1=\left( {8+\frac{{24i}}{n}+\frac{{18{{i}^{2}}}}{{{{n}^{2}}}}+\frac{{12i}}{n}+\frac{{36{{i}^{2}}}}{{{{n}^{2}}}}+\frac{{27{{i}^{3}}}}{{{{n}^{3}}}}} \right)-1\\&=\frac{{36i}}{n}+\frac{{54{{i}^{2}}}}{{{{n}^{2}}}}+\frac{{27{{i}^{3}}}}{{{{n}^{3}}}}+7\end{align}$
Width = $ \displaystyle \Delta x=\frac{{b-a}}{n}=\frac{{5-2}}{n}=\frac{3}{n}$.
Use the area formula: $ \displaystyle \begin{align}\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\text{height}\,\cdot \,\text{width}} \right)=}}\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left[ {\frac{{36i}}{n}+\frac{{54{{i}^{2}}}}{{{{n}^{2}}}}+\frac{{27{{i}^{3}}}}{{{{n}^{3}}}}+7} \right]}}\left( {\frac{3}{n}} \right)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\frac{{108i}}{{{{n}^{2}}}}+\frac{{162{{i}^{2}}}}{{{{n}^{3}}}}+\frac{{81{{i}^{3}}}}{{{{n}^{4}}}}+\frac{{21}}{n}} \right)}}\end{align}$
Now that we’ve simplified, separate the summations if needed, and bring out the numbers and $ n$’s to the outside of the summations (for the last term, we have to leave the summation with a “1”, which is a constant):
$ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\frac{{108i}}{{{{n}^{2}}}}+\frac{{162{{i}^{2}}}}{{{{n}^{3}}}}+\frac{{81{{i}^{3}}}}{{{{n}^{4}}}}+\frac{{21}}{n}} \right)}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{108}}{{{{n}^{2}}}}\,\sum\limits_{{i=1}}^{n}{i}+\frac{{162}}{{{{n}^{3}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{2}}+\frac{{81}}{{{{n}^{4}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{3}}+\frac{{21}}{n}\sum\limits_{{i=1}}^{n}{1}}}}}} \right)$
Remember the summation formulas:
$ \displaystyle \sum\limits_{{i=1}}^{n}{{c=cn\,\,(c\text{ is a constant)}\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{i=\frac{{n\left( {n+1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{2}}=\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{3}}=\frac{{{{n}^{2}}{{{\left( {n+1} \right)}}^{2}}}}{4}\,\,}}$
Now get rid of the $ \Sigma $’s by turning all the $ \Sigma i$’s into $ n$’s using the formulas above. Then simplify, and take the limits, remembering that $ \displaystyle \underset{{n\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{n}^{r}}}}=0$.
$ \displaystyle \begin{align}&\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{108}}{{{{n}^{2}}}}\,\sum\limits_{{i=1}}^{n}{i}+\frac{{162}}{{{{n}^{3}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{2}}+\frac{{81}}{{{{n}^{4}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{3}}+\frac{{21}}{n}\sum\limits_{{i=1}}^{n}{1}}}}}} \right)\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left[ {\frac{{108}}{{{{n}^{2}}}}\left( {\frac{{n\left( {n+1} \right)}}{2}} \right)+\frac{{162}}{{{{n}^{3}}}}\left( {\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}} \right)+\frac{{81}}{{{{n}^{4}}}}\left( {\frac{{{{n}^{2}}{{{\left( {n+1} \right)}}^{2}}}}{4}} \right)+\left( {\frac{{21}}{n}} \right)\left( {1n} \right)} \right]\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left[ {\frac{{{}^{{54}}\cancel{{108}}}}{{{}_{n}\cancel{{{{n}^{2}}}}}}\left( {\frac{{\cancel{n}\left( {n+1} \right)}}{{{{{\cancel{2}}}_{1}}}}} \right)+\frac{{{}^{{27}}\cancel{{162}}}}{{{}_{{{{n}^{2}}}}\cancel{{{{n}^{3}}}}}}\left( {\frac{{\cancel{n}\left( {n+1} \right)\left( {2n+1} \right)}}{{{{{\cancel{6}}}_{1}}}}} \right)+\frac{{81}}{{{}_{{{{n}^{2}}}}\cancel{{{{n}^{4}}}}}}\left( {\frac{{\cancel{{{{n}^{2}}}}{{{\left( {n+1} \right)}}^{2}}}}{4}} \right)+\left( {\frac{{21}}{{\cancel{n}}}} \right)\left( {1\cancel{n}} \right)} \right]\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{54n+54}}{n}+\frac{{54{{n}^{2}}+81n+27}}{{{{n}^{2}}}}+\frac{{81{{n}^{2}}+162n+81}}{{4{{n}^{2}}}}+21} \right)\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {54+{{{\cancel{{\frac{{54}}{n}}}}}^{0}}+54+{{{\cancel{{\frac{{81}}{n}+\frac{{27}}{{{{n}^{2}}}}}}}}^{0}}+\frac{{81}}{4}+{{{\cancel{{\frac{{162}}{{4n}}+\frac{{81}}{{4{{n}^{2}}}}}}}}^{0}}+21} \right)=54+54+\frac{{81}}{4}+21=149.25\end{align}$
Here’s a tricky problem on area by limit definition (sums) that you might see on a test. And (spoiler alert!), you’ll see in the possible answers how we’ll do these problems so much easier using Definite Integration in the next section.
Problem: $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\,\sum\limits_{{i=1}}^{n}{{\,\frac{2}{n}\left[ {{{{\left( {\frac{2}{n}} \right)}}^{2}}+{{{\left( {\frac{4}{n}} \right)}}^{2}}+{{{\left( {\frac{6}{n}} \right)}}^{2}}+…+{{{\left( {\frac{2n}{n}} \right)}}^{2}}} \right]}}$ = which one of the following areas?
a) Area of the region between $ \displaystyle y=\frac{1}{{{{x}^{2}}}}$ and the $ x$-axis from 0 and 2 (which is $ \displaystyle \int_{0}^{2}{{\frac{1}{{{{x}^{2}}}}\,}}dx$)
b) Area of the region between $ y={{x}^{2}}$ and the $ x$-axis from 0 and $ n$ (which is $ \displaystyle \int_{0}^{n}{{{{x}^{2}}\,}}dx$)
c) Area of the region between $ y={{x}^{2}}$ and the $ x$-axis from 0 and 2 (which is $ \displaystyle \int_{0}^{2}{{{{x}^{2}}\,}}dx$)
d) Area of the region between $ \displaystyle y=\frac{1}{{{{x}^{2}}}}$ and the $ \displaystyle x$-axis from 2 and 4 (which is $ \displaystyle \int_{2}^{4}{{\frac{1}{{{{x}^{2}}}}\,}}dx$)
Solution:
c) This is a limit sum with width $ \displaystyle \Delta x=\frac{{b-a}}{n}=\frac{{2-0}}{n}=\frac{2}{n}$, and it starts at $ x=0$ since we’re not adding anything to the $ \displaystyle \frac{2}{n}$. The height of each rectangles for $ y={{x}^{2}}$ is $ \displaystyle f\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)=f\left( {0+\frac{{\left( {2-0} \right)i}}{n}} \right)=f\left( {\frac{{2i}}{n}} \right)={{\left( {\frac{{2i}}{n}} \right)}^{2}}$, where $ i$ goes from 1 to $ n$; this is what’s shown above.
Understand these problems, and practice, practice, practice!
On to Definite Integration – you’re ready!