# 4.3: The Area Problem

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use sigma notation to evaluate sums of rectangular areas
• Find limits of upper and lower sums
• Use the limit definition of area to solve problems

## Introduction

In The Lesson The Calculus we introduced the area problem that we consider in integral calculus. The basic problem was this: \begin{align*}f(x) = x^2.\end{align*}

Suppose we are interested in finding the area between the \begin{align*}x-\end{align*}axis and the curve of \begin{align*}f(x) = x^2,\end{align*} from \begin{align*}x = 0\end{align*} to \begin{align*}x = 1.\end{align*}

We approximated the area by constructing four rectangles, with the height of each rectangle equal to the maximum value of the function in the sub-interval.

We then summed the areas of the rectangles as follows:

\begin{align*} R_1 & = \frac{1} {4} \cdot f \left ( \frac{1} {4} \right ) = \frac{1} {64}, \\ R_2 & = \frac{1} {4} \cdot f \left ( \frac{1} {2} \right ) = \frac{1} {16}, \\ R_3 & = \frac{1} {4} \cdot f \left ( \frac{3} {4} \right ) = \frac{9} {64}, \\ R_4 & = \frac{1} {4} \cdot f(1) = \frac{1} {4},\end{align*}

and \begin{align*}\ R_1 + R_2 + R_3 + R_4 = \frac{30} {64} = \frac{15} {32} \approx 0.46.\end{align*}

We call this the upper sum since it is based on taking the maximum value of the function within each sub-interval. We noted that as we used more rectangles, our area approximation became more accurate.

We would like to formalize this approach for both upper and lower sums. First we note that the lower sums of the area of the rectangles results in \begin{align*}R_1 + R_2 + R_3 + R_4 = 13/64 \approx 0.20\end{align*} Our intuition tells us that the true area lies somewhere between these two sums, or \begin{align*}0.20 < \;\mathrm{Area}\ < 0.46\end{align*} and that we will get closer to it by using more and more rectangles in our approximation scheme.

In order to formalize the use of sums to compute areas, we will need some additional notation and terminology.

Sigma Notation

In The Lesson The Calculus we used a notation to indicate the upper sum when we increased our rectangles to \begin{align*} N = 16\end{align*} and found that our approximation \begin{align*}A = \frac{1432} {4096} \approx .34\end{align*}. The notation we used to enabled us to indicate the sum without the need to write out all of the individual terms. We will make use of this notation as we develop more formal definitions of the area under the curve.

Let’s be more precise with the notation. For example, the quantity \begin{align*} A = \sum R_i\end{align*} was found by summing the areas of \begin{align*}N = 16\end{align*} rectangles. We want to indicate this process, and we can do so by providing indices to the symbols used as follows:

\begin{align*} A = \sum_{i=1}^{16} R_i = R_1 + R_2 + R_3 + . . . + R_{15} + R_{16}.\end{align*}

The sigma symbol with these indices tells us how the rectangles are labeled and how many terms are in the sum.

Useful Summation Formulas

We can use the notation to indicate useful formulas that we will have occasion to use. For example, you may recall that the sum of the first \begin{align*}n\end{align*} integers is \begin{align*}n( n + 1)/2.\end{align*} We can indicate this formula using sigma notation. The formula is given here along with two other formulas that will become useful to us.

\begin{align*}\sum_{i = 1}^n i & = \frac{n(n + 1)} {2}, \\ \sum_{i = 1}^n i^2\ & = \frac{n(n + 1) (2n + 1)} {6}, \\ \sum_{i = 1}^n i^3 & = \left [ \frac{n(n + 1)} {2} \right]^2.\end{align*}

We can show from associative, commutative, and distributive laws for real numbers that

\begin{align*}\sum_{i = 1}^n (a_i + b_i) = \sum_{i = 1}^n (a_i) + \sum_{i = 1}^n (b_i)\end{align*} and

\begin{align*}\sum_{i = 1}^n (k a_i) = k \sum_{i = 1}^n (a_i).\end{align*}

Example 1:

Compute the following quantity using the summation formulas:

\begin{align*} \sum_{i = 1}^{10} 2i(i - 6i).\end{align*}

Solution:

\begin{align*} \sum_{i = 1}^{10} 2i(i - 6) & = \sum_{i = 1}^{10} (2i^2 - 12i) = 2 \sum_{i = 1}^{10} i^2 - 12 \sum_{i = 1}^{10} i \\ & = 2 \left ( \frac{(10) (10 + 1) (2 \cdot 10 + 1)} {6} \right) - 12 \left ( \frac{ (10) (11)} {2} \right) \\ & = 770 - 660 = 110.\end{align*}

Another Look at Upper and Lower Sums

We are now ready to formalize our initial ideas about upper and lower sums.

Let \begin{align*}f\end{align*} be a bounded function in a closed interval \begin{align*}[a, b]\end{align*} and \begin{align*}P = [x_0, \ldots, x_n]\end{align*} the partition of \begin{align*}[a , b]\end{align*} into \begin{align*}n\end{align*} subintervals.

We can then define the lower and upper sums, respectively, over partition \begin{align*}P\end{align*}, by

\begin{align*} S(P) & = \sum_{1}^n m_i (x_i - x_{i - 1}) = m_1 (x_1 - x_0) + m_2 (x_2 - x_1) + \ldots+ m_n (x_n - x_{n - 1}), \\ T(P) & = \sum_{1}^n M_i (x_i - x_{i - 1}) = M_1 (x_1 - x_0) + M_2 (x_2 - x_1) + \ldots+ M_n (x_n - x_{n - 1}).\end{align*}

where \begin{align*}m_i\end{align*} is the minimum value of \begin{align*}f\end{align*} in the interval of length \begin{align*}x_i - x_{i - 1}\end{align*} and \begin{align*}M_i\end{align*} is the maximum value of \begin{align*}f\end{align*} in the interval of length \begin{align*}x_i - x_{i - 1}.\end{align*}

The following example shows how we can use these to find the area.

Example 2:

Show that the upper and lower sums for the function \begin{align*}f(x) = x^2,\end{align*} from \begin{align*}x = 0\end{align*} to \begin{align*}x = 1,\end{align*} approach the value \begin{align*}A = 1/3.\end{align*}

Solution:

Let \begin{align*}P\end{align*} be a partition of \begin{align*}n\end{align*} equal sub intervals over \begin{align*}[0, 1].\end{align*} We will show the result for the upper sums. By our definition we have

\begin{align*} T(P) = \sum_{1}^n M_i (x_i - x_{i - 1}) = M_1 (x_1 - x_0) + M_2 (x_2 - x_1) + \ldots + M_n (x_n - x_{n - 1}).\end{align*}

We note that each rectangle will have width \begin{align*} \frac{1} {n},\end{align*} and lengths \begin{align*} \left (\frac{1} {n} \right)^2 , \left (\frac{2} {n} \right)^2 , \left (\frac{3} {n} \right)^2 , \ldots, \left (\frac{n} {n} \right)^2\end{align*} as indicated:

\begin{align*} T(P) & = \sum_{1}^n M_i (x_i - x_{i - 1}) = M_1 (x_1 - x_0) + M_2 (x_2 - x_1) + \ldots + M_n (x_n - x_{n - 1}) \\ & = \frac{1} {n} \left( \frac{1} {n} \right)^2 + \frac{1} {n} \left( \frac{2} {n} \right)^2 + \frac{1} {n} \left( \frac{3} {n} \right)^2 + \ldots + \frac{1} {n} \left( \frac{n} {n} \right)^2 \\ & = \frac{1} {n} \left( \frac{1} {n} \right)^2 ( 1^2 + 2^2 + 3^2 + \ldots + n^2) \\ & = \left ( \frac{1} {n^3} \right) (1^2 + 2^2 + 3^2 + \ldots + n^2) = \left ( \frac{1} {n^3} \right) \left ( \frac{n(n +1) (2n + 1)} {6} \right) = \left ( \frac{(n + 1) (2n + 1)} {6n^2} \right).\end{align*}

We can re-write this result as:

\begin{align*} \frac{(n + 1)(2n + 1)} {6n^2} = \frac{1} {6} \left ( \frac{n + 1} {n} \right) \left ( \frac{2n + 1} {n} \right) = \frac{1} {6} \left (1 + \frac{1} {n} \right) \left (2 + \frac{1} {n} \right).\end{align*}

We observe that as

\begin{align*} n \rightarrow +\infty, \frac{1} {6} \left ( 1 + \frac{1} {n} \right) \left (2 + \frac{1} {n} \right) \rightarrow \frac{1} {3}.\end{align*}

We now are able to define the area under a curve as a limit.

Definition
Let \begin{align*}f\end{align*} be a continuous function on a closed interval \begin{align*}[a ,b].\end{align*} Let \begin{align*}P\end{align*} be a partition of \begin{align*}n\end{align*} equal sub intervals over \begin{align*}[a ,b].\end{align*} Then the area under the curve of \begin{align*}f\end{align*} is the limit of the upper and lower sums, that is

\begin{align*} A = \lim_{n \rightarrow +\infty} S(P) = \lim_{n \rightarrow +\infty} T(P).\end{align*}

Example 3:

Use the limit definition of area to find the area under the function \begin{align*}f(x) = 4 - x\end{align*} from \begin{align*}1\end{align*} to \begin{align*}x = 3.\end{align*}

Solution:

If we partition the interval \begin{align*}[1 ,3]\end{align*} into \begin{align*}n\end{align*} equal sub-intervals, then each sub-interval will have length \begin{align*} \frac{3 - 1} {n} = \frac{2} {n}\end{align*} and height \begin{align*}3 - i \triangle x\end{align*} as \begin{align*}i\end{align*} varies from \begin{align*}1\end{align*} to \begin{align*}n.\end{align*} So we have \begin{align*} \triangle x = \frac{2} {n}\end{align*} and

\begin{align*}S(P) &= \sum_1^n (3 - i \triangle x) \triangle x = \sum_1^n (3 \triangle x) - \sum_1^n i(\triangle x )^2\\ &= (3 \triangle x)n - \frac{n(n + 1)} {2} (\triangle x)^2.\end{align*}

Since \begin{align*} \triangle x = \frac{2} {n}\end{align*} , we then have by substitution

\begin{align*} (3 \triangle x)n - \frac{n(n + 1)} {2} ( \triangle x)^2 = 6 - \left (2 + \frac{2} {n} \right) = 4 + \frac{2} {n} \rightarrow 4\end{align*} as \begin{align*} n \rightarrow \infty\end{align*} . Hence the area is \begin{align*}A = 4.\end{align*}

This example may also be solved with simple geometry. It is left to the reader to confirm that the two methods yield the same area.

## Lesson Summary

1. We used sigma notation to evaluate sums of rectangular areas.
2. We found limits of upper and lower sums.
3. We used the limit definition of area to solve problems.

## Review Questions

In problems #1–2 , find the summations.

1. \begin{align*} \sum_{i = 1}^{10} i(2i - 3)\end{align*}
2. \begin{align*} \sum_{i = 1}^n (3 - i) (2 + i)\end{align*}

In problems #3–5, find \begin{align*}S(P)\end{align*} and \begin{align*}T(P)\end{align*} under the partition \begin{align*}P.\end{align*}

1. \begin{align*} f(x) = 1 - x^2,\end{align*} \begin{align*}P = \left \{ 0, \frac{1} {2}, 1, \frac{3} {2}, 2 \right \}\end{align*}
2. \begin{align*} f(x) = 2x^2,\end{align*} \begin{align*}P = \left \{ -1, -\frac{1} {2}, 0, \frac{1} {2}, 1 \right \}\end{align*}
3. \begin{align*} f(x) = \frac{1} {x},\end{align*} \begin{align*}P = \left \{ -4, -3, -2, -1 \right \}\end{align*}

In problems #6–8, find the area under the curve using the limit definition of area.

1. \begin{align*}f(x) = 3 x + 5\end{align*} from \begin{align*}x = 2\end{align*} to \begin{align*}x = 6.\end{align*}
2. \begin{align*}f(x) = x^2\end{align*} from \begin{align*}x = 1\end{align*} to \begin{align*}x = 3.\end{align*}
3. \begin{align*} f(x) = \frac{1} {x}\end{align*} from \begin{align*}x = 1\end{align*} to \begin{align*}x = 4.\end{align*}

In problems #9–10, state whether the function is integrable in the given interval. Give a reason for your answer.

1. \begin{align*}f(x) = |x - 2|\end{align*} on the interval \begin{align*}[1 ,4]\end{align*}
2. \begin{align*} f(x) = \begin{cases} 1 & \;\mathrm{if}\ x\;\mathrm{ is\ rational}\\ -1 & \;\mathrm {if}\ x\;\mathrm{ is\ irrational} \end{cases}\end{align*} on the interval \begin{align*}[0 ,1]\end{align*}

1. \begin{align*} \sum_{i = 1}^{10} i(2i - 3) = 605\end{align*}
2. \begin{align*} \sum_{i = 1}^n (3 - i) (2 + i) = \frac{n} {3} (19 - n^2)\end{align*}
3. \begin{align*}S (P) = -1.75\end{align*}; \begin{align*}T(P) = 0.25\end{align*} (note that we have included areas under the \begin{align*}x-\end{align*}axis as negative values.)
4. \begin{align*}S (P) = 0.5\end{align*}; \begin{align*}T(P) = 2.5\end{align*}
5. \begin{align*}S (P) = -1.83\end{align*}; \begin{align*}T(P) = -1.08\end{align*}
6. \begin{align*}\;\mathrm{Area} = 68\end{align*}
7. \begin{align*}\;\mathrm{Area} = \frac{26} {3}\end{align*}
8. \begin{align*}\;\mathrm{Area} = \frac{15} {16}\end{align*}
9. Yes, since \begin{align*}f(x) = |x - 2|\end{align*} is continuous on \begin{align*}[1, 4]\end{align*}
10. No, since \begin{align*}S(P) = -1\end{align*}; \begin{align*}T(P) = 1\end{align*}

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