# Riemann Sums and Area by Limit Definition

This section covers:

# Introduction to Riemann Sums

I’m convinced the reason they teach you Riemann Sums 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.

Riemann Sums can be used to approximate the area under curves, which will be acquired much easier by just taking the integral of the function between two different $$x$$ values (we’ll do this in the Definite Integral section). But, alas, we have to learn these more difficult methods first.

Let’s say we wanted to get the area of the region between $$x=1$$ and $$x=2$$ for certain “curved” functions. 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? Do you see how, for some functions, the “right hand” sums (where the right hand sides of the rectangles touch the function) over-approximates the area (upper sum), and for others, it under-approximates the area (lower sum)? 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 under a curve, 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. Let’s use the example of  $$y={{x}^{2}}$$  between $$x=1$$ and $$x=2$$, with $$n=8$$ (number of intervals).

Using Lower Sum to Approximate Area between x = 1 and x = 2:

Function $$y={{x}^{2}}$$, $$n=8$$, left-hand sum:

Note that for interval $$[a,b]$$, to get the width ($$\Delta x$$) of each rectangle, we can subtract the small number $$(a)$$ from the larger number $$(a)$$ 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, we need to 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 $$y$$’s, we need to 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 $$y$$ 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( {1\frac{1}{8}} \right)}}^{2}}+{{{\left( {1\frac{2}{8}} \right)}}^{2}}+{{{\left( {1\frac{3}{8}} \right)}}^{2}}+{{{\left( {1\frac{4}{8}} \right)}}^{2}}+{{{\left( {1\frac{5}{8}} \right)}}^{2}}+{{{\left( {1\frac{6}{8}} \right)}}^{2}}+{{{\left( {1\frac{7}{8}} \right)}}^{2}}} \right)\\&=\left( {\frac{1}{8}} \right)17.1875\approx 2.148\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 sums:

<
 Lower, Upper and Midpoint Sum Problems Solution Use the 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.   Note: Actual area = 2.667 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)\\&=\left( {\frac{1}{4}} \right)10.073=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)=\left( {\frac{1}{4}} \right)11.073\\&=2.768\end{align} 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 = 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:   $$\displaystyle \begin{array}{l}\text{area}\approx 1\cdot f\left( {\frac{3}{2}} \right)+1\cdot f\left( {\frac{5}{2}} \right)+1\cdot f\left( {\frac{7}{2}} \right)+1\cdot f\left( {\frac{9}{2}} \right)\\=1\left[ {\left( {{{{\left( {\frac{3}{2}} \right)}}^{2}}+2\left( {\frac{3}{2}} \right)} \right)+\left( {{{{\left( {\frac{5}{2}} \right)}}^{2}}+2\left( {\frac{5}{2}} \right)} \right)+\left( {{{{\left( {\frac{7}{2}} \right)}}^{2}}+2\left( {\frac{7}{2}} \right)} \right)+\left( {{{{\left( {\frac{9}{2}} \right)}}^{2}}+2\left( {\frac{9}{2}} \right)} \right)} \right]\\=65\end{array}$$

# Trapezoidal Rule

The Trapezoidal Rule approximates area, but uses trapezoids instead of rectangles. Remember that a trapezoid is a quadrilateral (four-sided figure) where the bases are parallel and the other two sides aren’t.

Remember from Geometry how we 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 looks like; notice how the trapezoids are turned on their sides.

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. When we do this, we come up with the definition of the Trapezoidal Rule:

The Trapezoidal Rule:

Let $$f$$ be continuous on interval $$\left[ {a,\,b} \right]$$. We can approximate $$\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$ 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 $$\displaystyle \frac{{f\left( {x{}_{1}} \right)+f\left( {x{}_{2}} \right)}}{2},\frac{{f\left( {x{}_{2}} \right)+f\left( {x3} \right)}}{2}$$ 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.

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 some problems:

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 0 100 200 300 400 500 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.

Total Area =      \displaystyle \begin{align}&\frac{{0+150}}{2}\cdot 100+\frac{{150+130}}{2}\cdot 100+\frac{{130+140}}{2}\cdot 100+\frac{{140+160}}{2}\cdot 100+\frac{{160+0}}{2}\cdot 100\\\,&=\frac{{100}}{2}\left( {150+280+270+300+160} \right)=50\cdot 1160=58000\end{align}

Multiply this by the depth of the pond (25 feet) to get Volume = 1,450,000 feet.

Note that we could have also used this Trapezoidal Rule equation to get the area, since the distances from the west end of the pond are all 100 feet apart:

\displaystyle \begin{align}A&=\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{{500-0}}{{2\left( 5 \right)}}\left( {0+2\left( {150} \right)+2\left( {130} \right)+2\left( {140} \right)+2\left( {160} \right)+0} \right)=58000\end{align}

Here are a few more 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:   $$\int\limits_{0}^{1}{{\sqrt{{2+{{x}^{3}}}}}}\,dx$$   . Divide 0 to 1 into 4 intervals, so $$f\left( x \right)$$’s will be 0, .25, .5, .75, 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)\\&=\frac{1}{8}\left( {12.01} \right)\\&\approx 1.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(Actual value}\approx \text{ 1}\text{.497)}\end{align} Using the trapezoidal rule with 3 equal subintervals, approximate the integral:   $$\int\limits_{{\frac{\pi }{2}}}^{\pi }{{\sqrt{{2x}}}}\sin \left( x \right)\,dx$$ Divide $$\displaystyle \frac{\pi }{2}$$ to $$\pi$$ into 3 intervals, so $$f\left( x \right)$$’s will be $$\displaystyle \frac{\pi }{2},\,\,\frac{{4\pi }}{6},\,\,\frac{{5\pi }}{6},$$ and $$\pi$$.   (To get these, subtract $$\pi$$ from $$\displaystyle \frac{\pi }{2}$$ to get $$\displaystyle \frac{\pi }{2}$$, and then divide by 3 to get $$\displaystyle \frac{\pi }{6}$$. Then add $$\displaystyle \frac{\pi }{6},\,\,\frac{{2\pi }}{6},$$ and $$\displaystyle \frac{{3\pi }}{6}$$ to the starting point $$\displaystyle \frac{\pi }{2}$$).   \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{{4\pi }}{6}} \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( {1.7725+2\cdot 1.7725+2\cdot 1.441+0} \right)\\&\approx 2.1466\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(Actual value }\approx \text{ 2}\text{.0617)}\end{align}

# 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 the sigma sign $$\Sigma$$ 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 $$y$$-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 $$\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$.

Here are the formulas you’ll really have to use:

$$\displaystyle \begin{array}{l}\text{Area between (}x=a\text{ and }x=b)\,\,\text{for function }f=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)\cdot \left( {\frac{{\left( {b-a} \right)}}{n}} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Height}\,\,\,\,\,\,\,\,\,\,\,\text{Width}\end{array}$$   (“memorize” this one!)

To simplify and get rid of summation signs, use these summation formulas (usually given):

$$\displaystyle \sum\limits_{{i=1}}^{n}{{c=cn\,\,(c\text{ is a constant)}\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{i=\frac{{n\left( {n+1} \right)}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{2}}=\frac{{n\left( {n+1} \right)\left( {2n+1} \right)}}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\sum\limits_{{i=1}}^{n}{{{{i}^{3}}=\frac{{{{n}^{2}}{{{\left( {n+1} \right)}}^{2}}}}{4}\,\,}}$$

What you want to do is use the area formula with the given function and interval, 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). Then use the above summation formulas to turn “$$i$$’s” into “$$n$$’s”. Then simplify and take the limit of what’s left (remember that $$\displaystyle \underset{{n\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{n}^{r}}}}=0$$) .  I know it’s tedious, but it works!Let’s show an example to see how this works:

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, we have to leave the summation with a “1”, which is a constant):

$$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{\left( {\frac{{8{{i}^{2}}}}{{{{n}^{3}}}}+\frac{8}{n}} \right)}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\sum\limits_{{i=1}}^{n}{{\frac{{8{{i}^{2}}}}{{{{n}^{3}}}}+}}\sum\limits_{{i=1}}^{n}{{\frac{8}{n}}}} \right)=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{8}{{{{n}^{3}}}}\,\sum\limits_{{i=1}}^{n}{{{{i}^{2}}}}+\frac{8}{n}\,\sum\limits_{{i=1}}^{n}{1}} \right)$$

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 \require {cancel} \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( {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 Integrals in the next section.

Problem:

$$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\,\sum\limits_{{i=1}}^{n}{{\,\frac{1}{n}\left[ {{{{\left( {\frac{1}{n}} \right)}}^{2}}+{{{\left( {\frac{2}{n}} \right)}}^{2}}+{{{\left( {\frac{3}{n}} \right)}}^{2}}+…+{{{\left( {\frac{n}{n}} \right)}}^{2}}} \right]}}$$ =  which one of the following areas?

a) Area of the region between $$y=x$$ and the $$x$$-axis from 0 and 1 (which is $$\displaystyle \int_{0}^{1}{{x\,}}dx$$)

b) Area of the region between $$\displaystyle y=\frac{1}{{{{x}^{2}}}}$$ and the $$x$$-axis from 0 and 1 (which is $$\displaystyle \int_{0}^{1}{{\frac{1}{{{{x}^{2}}}}\,}}dx$$)

c) Area of the region $$y={{x}^{2}}$$ between and the $$x$$-axis from 0 and 1 (which is $$\displaystyle \int_{0}^{1}{{{{x}^{2}}\,}}dx$$)

d) Area of the region between $$y={{x}^{2}}$$ and the $$\displaystyle x$$-axis from 1 and 2 (which is $$\displaystyle \int_{1}^{2}{{{{x}^{2}}\,}}dx$$)

Solution:

c) This is a limit sum with width $$\displaystyle \Delta x=\frac{{b-a}}{n}=\frac{{1-0}}{n}=\frac{1}{n}$$, and it starts at $$x=0$$ since we’re not adding anything to the $$\displaystyle \frac{1}{n}$$. The height of each rectangles is $$\displaystyle \frac{1}{n}$$, and we are taking the square of it, so we have the area between the function $$y={{x}^{2}}$$ and the $$x$$-axis between 0 and 1.

Understand these problems, and practice, practice, practice!

On to Definite Integration – you’re ready!

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