Definition of the Definite Integral

A Riemann sum approximates the area between a curve and the $x$ -axis by a collection of rectangles. The definite integral is what those approximations converge to as the rectangles become infinitely thin.

Formula	Name	Notes
$S_{n} = \sum_{k = 1}^{n} f (x_{k}^{*}) Δ_{k} x$	Riemann Sum	$Δ_{k} x = x_{k} - x_{k - 1}$ , and $x_{k}^{*}$ is any point in the $k$ -th subinterval
$Δ x = \frac{b - a}{n}$	Uniform Subinterval Width	Used when the $n$ subintervals all have the same width
$x_{k}^{*} = x_{k - 1}$	Left Riemann Sum	Sample point at the left endpoint
$x_{k}^{*} = x_{k}$	Right Riemann Sum	Sample point at the right endpoint
$x_{k}^{*} = \frac{x_{k - 1} + x_{k}}{2}$	Middle Riemann Sum	Sample point at the midpoint
$\int_{a}^{b} f (x) d x = lim_{max Δ_{k} x \to 0} \sum_{k = 1}^{n} f (x_{k}^{*}) Δ_{k} x$	Definite Integral	Exists whenever $f$ is continuous on $[a, b]$
$\int_{a}^{b} f = - \int_{b}^{a} f$	Reversal Rule	Swapping the limits reverses the sign
$\int_{a}^{a} f (x) d x = 0$	Zero-Width Rule	An interval of zero length contributes nothing
$\sum_{k = 1}^{n} 1 = n, \sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$	Summation Formulas	Needed to evaluate Riemann sums in closed form
$\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$	Sum of Squares	Needed for integrands involving $x^{2}$

Riemann Sums

Let $f (x)$ be defined for $a \leq x \leq b$ . Choose $n - 1$ points arbitrarily between a and b

a = x_{0} < x_{1} < x_{2} < \dots < x_{n - 1} < x_{n} = b,

thus dividing the interval $I = [a, b]$ into $n$ subintervals:

[a, x_{1}], [x_{1}, x_{2}], \dots, [x_{n - 1}, b] .

In each of these subintervals, take a value $x = x_{k}^{*}$ where $x_{k - 1} \leq x_{k}^{*} \leq x_{k}$ , and form the sum

S_{n}=f(x_{1}^{*})(x_{1}-x_{0})+f(x_{2}^{*})(x_{2}-x_{1})+\cdots+f(x_{n}^{*})(x_{n}-x_{n-1}). \tag{1}

Definition 1.

The sum in Equation (1) is called a Riemann sum for $f$ on the interval $[a, b]$ , after the German mathematician Georg Friedrich Bernhard Riemann (1826-1866). The points $x_{k}^{*}$ are called sample points.

Now let us see what the geometrical interpretation of $S_{n}$ is. First suppose $f (x) \geq 0$ for all $x \in [a, b]$ . In this case each term of the sum is the area of a rectangle whose base is a subinterval and whose height is the value of $f$ at the sample point:

$f (x_{1}^{*}) (x_{1} - x_{0}) =$ the area of the rectangle $A P_{1} R_{1} M_{1}$ ,

$f (x_{2}^{*}) (x_{2} - x_{1}) =$ the area of the rectangle $M_{1} P_{2} R_{2} M_{2}$ ,

$⋮$

$f (x_{n - 1}^{*}) (x_{n - 1} - x_{n - 2}) =$ the area of the rectangle $M_{n - 2} P_{n - 1} R_{n - 1} M_{n - 1}$ ,

$f (x_{n}^{*}) (x_{n} - x_{n - 1}) =$ the area of the rectangle $M_{n - 1} P_{n} R_{n} B$ .

TikZ figure — The rectangles of a Riemann sum for $f$ on $[a, b]$ .

Thus $S_{n}$ represents the total area of all the rectangles in the figure, and approximates the area of the region that lies between the curve $y = f (x)$ and the $x$ -axis between $x = a$ and $x = b$ .

If $f (x)$ takes both positive and negative values on $[a, b]$ , then $S_{n}$ is an approximation for the net area (or the signed area, or sometimes the "net signed area"); that is, the area below the curve $y = f (x)$ and above the $x$ -axis minus the area above the curve $y = f (x)$ and below the $x$ -axis from $x = a$ to $x = b$ .

The Riemann sum (1) is often shortened as

S_{n} = \sum_{k = 1}^{n} f (x_{k}^{*}) (x_{k} - x_{k - 1}),

where $Σ$ (sigma), the Greek form of the letter S, stands for the word "summation." The whole expression indicates that the sum is to be taken of all terms obtained from $f (x_{k}^{*}) (x_{k} - x_{k - 1})$ by giving to k in succession the values $1, 2, 3, \dots, n$ .

By introducing the symbol

Δ_{k} x = x_{k} - x_{k - 1},

we can shorten it further and write:

S_{n}=\sum_{k=1}^{n}f(x_{k}^{*})\,\Delta_{k}x. \tag{2}

Please note that here $Δ_{k}$ is not a factor. It denotes a difference, in this case the length of the $k$ -th subinterval. Many books write $Δ x_{k}$ instead of $Δ_{k} x$ ; the two notations mean the same thing.

Some Specific Types of Riemann Sums

The definition allows the sample point $x_{k}^{*}$ to be any point of the $k$ -th subinterval. Certain standard choices have their own names.

Definition 2.

If the sample point in each subinterval $x_{k}^{*}$ is the left endpoint of the subinterval, that is, if $x_{k}^{*} = x_{k - 1}$ for all k, then $S_{n}$ is called a left Riemann sum.

Left Riemann sum with rectangle heights taken at the left endpoint of each subinterval — A left Riemann sum.

Definition 3.

If the sample point in each subinterval $x_{k}^{*}$ is the right endpoint of the subinterval, that is, if $x_{k}^{*} = x_{k}$ for all k, then $S_{n}$ is called a right Riemann sum.

Right Riemann sum with rectangle heights taken at the right endpoint of each subinterval — A right Riemann sum.

Definition 4.

If the sample point in each subinterval $x_{k}^{*}$ is the midpoint of the subinterval, that is, if $x_{k}^{*} = (x_{k} + x_{k - 1}) / 2$ for all k, then $S_{n}$ is called a middle Riemann sum (or midpoint Riemann sum).

Middle Riemann sum with rectangle heights taken at the midpoint of each subinterval — A middle Riemann sum.

Example 1.

The graph of $f$ is shown below. Estimate the net area between the graph of $f$ and the $x$ -axis on the interval $[- 0.5, 4.5]$ using five subintervals of equal width and

(a) a left Riemann sum

(b) a right Riemann sum

(c) a midpoint Riemann sum.

Graph of a function f on the interval from negative 0.5 to 4.5 used to estimate net area — The graph of $f$ on $[- 0.5, 4.5]$ .

Solution

With $n = 5$ , $b = 4.5$ and $a = - 0.5$ , the width of the rectangles is

Δ x = \frac{b - a}{n} = \frac{4.5 - (- 0.5)}{5} = 1.

(a) The height of each rectangle in a left Riemann sum is the value of the function at the left boundary of each subinterval. The sum of the areas of the rectangles is

\begin{aligned} \Delta x&\left[f(-0.5)+f(0.5)+f(1.5)+f(2.5)+f(3.5)\right]\\ &=1.0\left[1.5+3+4+2.5+(-0.5)\right]=10.5. \end{aligned}

Left Riemann sum rectangles drawn on the graph of f over five subintervals — The left Riemann sum with five subintervals.

(b) The height of each rectangle in a right Riemann sum is the value of the function at the right boundary of each subinterval. The sum of the areas of the rectangles is

\begin{aligned} \Delta x&\left[f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)\right]\\ &=1.0\left[3+4+2.5+(-0.5)+(-0.5)\right]=8.5. \end{aligned}

Right Riemann sum rectangles drawn on the graph of f over five subintervals — The right Riemann sum with five subintervals.

(c) The height of each rectangle in a midpoint Riemann sum is the value of the function at the middle of each subinterval. There are five subintervals and we choose $x_{k}^{*}$ to be the midpoint of the $k$ -th one:

In the first subinterval $[- 0.5, 0.5]$ , $x_{1}^{*} = 0$ .
In the second subinterval $[0.5, 1.5]$ , $x_{2}^{*} = 1$ .
In the third subinterval $[1.5, 2.5]$ , $x_{3}^{*} = 2$ .
In the fourth subinterval $[2.5, 3.5]$ , $x_{4}^{*} = 3$ .
In the last subinterval $[3.5, 4.5]$ , $x_{5}^{*} = 4$ .

Therefore, the sum of the areas of the rectangles is

\begin{aligned} \sum_{k=1}^{5}f(x_{k}^{*})\Delta x &=\Delta x\left[f(0)+f(1)+f(2)+f(3)+f(4)\right]\\ &=1.0\left[2+3.5+3.5+1+(-1)\right]=9. \end{aligned}

Midpoint Riemann sum rectangles drawn on the graph of f over five subintervals — The midpoint Riemann sum with five subintervals.

Two further choices of sample point are important because they bracket every other Riemann sum.

Definition 5.

If the sample point in each subinterval $x_{k}^{*}$ is chosen such that $f (x_{k}^{*})$ is the maximum value of $f$ on the subinterval $[x_{k - 1}, x_{k}]$ for all k, then $S_{n}$ is called an upper Riemann sum.

Upper Riemann sum with each rectangle height equal to the maximum value of f on its subinterval — An upper Riemann sum.

Definition 6.

If the sample point in each subinterval $x_{k}^{*}$ is chosen such that $f (x_{k}^{*})$ is the minimum value of $f$ on the subinterval $[x_{k - 1}, x_{k}]$ for all k, then $S_{n}$ is called a lower Riemann sum.

Lower Riemann sum with each rectangle height equal to the minimum value of f on its subinterval — A lower Riemann sum.

Because every sample value lies between the minimum and the maximum on its subinterval, every Riemann sum for a given subdivision lies between the lower sum and the upper sum for that subdivision. In the example below, the five-rectangle estimates 10.5, 8.5 and 9 all lie between 5.5 and 13.

Example 2.

The graph of $f$ is the same as in the previous example. Estimate the net area between the graph of $f$ and the $x$ -axis on the interval $[- 0.5, 4.5]$ using five subintervals of equal width and

(a) an upper Riemann sum

(b) a lower Riemann sum.

Graph of the function f used for the upper and lower Riemann sum estimates — The graph of $f$ on $[- 0.5, 4.5]$ .

Solution

(a) As in the previous example, the width of the rectangles is

Δ x = \frac{b - a}{n} = \frac{4.5 - (- 0.5)}{5} = 1.

There are five subintervals: $[- 0.5, 0.5]$ , $[0.5, 1.5]$ , $[1.5, 2.5]$ , $[2.5, 3.5]$ , and $[3.5, 4.5]$ . The height of each rectangle is the maximum value of the function (the maximum $y$ -value) in the corresponding subinterval.

When $- 0.5 \leq x \leq 0.5$ , the maximum value of $f$ occurs at $x_{1}^{*} = 0.5$ .
When $0.5 \leq x \leq 1.5$ , the maximum value of $f$ occurs at $x_{2}^{*} = 1.5$ .
When $1.5 \leq x \leq 2.5$ , the maximum value of $f$ occurs at $x_{3}^{*} = 1.5$ .
When $2.5 \leq x \leq 3.5$ , the maximum value of $f$ occurs at $x_{4}^{*} = 2.5$ .
When $3.5 \leq x \leq 4.5$ , the maximum value of $f$ occurs at $x_{5}^{*} = 3.5$ or $x_{5}^{*} = 4.5$ .

Therefore, the upper Riemann sum is

\begin{aligned} \sum_{k=1}^{5}f(x_{k}^{*})\Delta x &=\Delta x\left[f(0.5)+f(1.5)+f(1.5)+f(2.5)+f(4.5)\right]\\ &=1.0\left[3+4+4+2.5+(-0.5)\right]=13. \end{aligned}

(b) The height of each rectangle is now the minimum value of the function (the minimum $y$ -value) in the corresponding subinterval.

When $- 0.5 \leq x \leq 0.5$ , the minimum value of $f$ occurs at $x_{1}^{*} = - 0.5$ .
When $0.5 \leq x \leq 1.5$ , the minimum value of $f$ occurs at $x_{2}^{*} = 0.5$ .
When $1.5 \leq x \leq 2.5$ , the minimum value of $f$ occurs at $x_{3}^{*} = 2.5$ .
When $2.5 \leq x \leq 3.5$ , the minimum value of $f$ occurs at $x_{4}^{*} = 3.5$ .
When $3.5 \leq x \leq 4.5$ , the minimum value of $f$ occurs at $x_{5}^{*} = 4$ .

Therefore, the lower Riemann sum is

\begin{aligned} \sum_{k=1}^{5}f(x_{k}^{*})\Delta x &=\Delta x\left[f(-0.5)+f(0.5)+f(2.5)+f(3.5)+f(4)\right]\\ &=1.0\left[1.5+3+2.5+(-0.5)+(-1)\right]=5.5. \end{aligned}

Limit of the Riemann Sums as $n \to \infty$

If the number of subintervals n increases so that the subinterval lengths (or widths) get smaller and the rectangles become thinner, then the Riemann sum $S_{n}$ becomes a better approximation to the net area between the curve $y = f (x)$ and the $x$ -axis.

Riemann sum with few rectangles giving a coarse approximation of the area — As the number of subintervals $n$ increases, $S_{n}$ becomes a better approximation to the net area between the curve $y = f (x)$ and the $x$ -axis.

Riemann sum with more rectangles giving a closer approximation of the area — A finer subdivision of the same interval.

Riemann sum with many thin rectangles nearly filling the region under the curve — With many thin rectangles, the Riemann sum is nearly indistinguishable from the region itself.

Now let the number of subintervals n increase indefinitely $(n \to \infty)$ , while the width of the widest subinterval approaches zero. If the sum (1), equivalently (2), approaches a number which is independent of the choice of the $x_{k}$ 's and of the $x_{k}^{*}$ 's, its limit is called the definite integral of $f (x)$ between a and b, and is denoted by $\int_{a}^{b} f (x) d x$ .

Definition 7.

The limit of the sum (2), provided that the number of subintervals n tends to infinity and at the same time the length of each subinterval approaches zero, is called the definite integral of $f (x)$ from a to b and is denoted by $\int_{a}^{b} f (x) d x$ . If this limit exists, the function is called integrable on the interval $[a, b]$ .

\int_{a}^{b} f (x) d x = lim_{max Δ_{k} x \to 0} \sum_{k = 1}^{n} f (x_{k}^{*}) Δ_{k} x .

In words: chop the interval up, build the rectangles, add their signed areas, and then squeeze all the widths down to zero. If that process settles on one number no matter how you chopped or where you sampled, that number is the integral.

The parts of the definite integral notation: integral sign, lower limit, upper limit, integrand and differential — The anatomy of the notation $\int_{a}^{b} f (x) d x$ .

We note that:

The use of the word "integral" and of the symbol $\int$ suggests a connection with the indefinite integrals of the previous chapter. This connection is established in Section 7.3.
The symbol $\int$ is a modified form of S, which stands for summation.
We call a the lower limit and b the upper limit of integration (or of the interval), and the function $f (x)$ the integrand.
If the integral $\int_{a}^{b} f (x) d x$ exists, $f (x)$ is called integrable between a and b.
If $f$ is continuous on the interval $(a, b)$ then $\int_{a}^{b} f (x) d x$ exists. The proof of this theorem is beyond the scope of an elementary book.

Dummy Variables

The letter k chosen for the summation index in the Riemann sum $S_{n} = \sum_{k = 1}^{n} f (x_{k}^{*}) Δ_{k} x$ is a dummy variable; that is, it can be replaced by any other letter such as i without affecting the value of the sum, provided of course that it is replaced everywhere k occurs:

S_{n} = \sum_{k = 1}^{n} f (x_{k}^{*}) Δ_{k} x = \sum_{i = 1}^{n} f (x_{i}^{*}) Δ_{i} x .

Similarly, the letter x chosen for the variable of integration in $\int_{a}^{b} f (x) d x$ is a dummy variable, which disappears in the final result. In other words, it can be replaced by any other letter we wish without affecting the integral:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (u) d u .

For example, if

\int_{1}^{2} f (x) d x = 32,

then

\int_{1}^{2} f (u) d u = 32.

Therefore, instead of $\int_{a}^{b} f (x) d x$ , we may sometimes simply write

\int_{a}^{b} f .

Unlike the case of the definite integral, the variable of integration in an indefinite integral is not a dummy variable, because it appears in the final result. For example,

\int x d x = \frac{1}{2} x^{2} + C_{1}, \int u d u = \frac{1}{2} u^{2} + C_{2},

and in this sense

\int x d x \neq \int u d u .

Reversing the Limits of Integration

In the previous discussion, we assumed $a < b$ . However, if we omit this condition and assume $b < a$ , we can still retain our definition of the integral; the only change is that we have to number the subdivision points from the right side, which makes the differences $Δ_{k} x$ negative. We are thus led to the relation

\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x .

In conformity with this, we define

\int_{a}^{a} f (x) d x = 0.

Definition 8.

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x and \int_{a}^{a} f (x) d x = 0.

  
    
      ∫
      
        a
      
      
        b
      
    
    f
    (
    x
    )
    
    d
    x
    =
  
 (area above the 
  x
-axis) 
  −
 (area below the 
  x
-axis), when 
  a
  <
  b
.

Evaluating Integrals from the Definition

The next two examples carry out the full limit computation. They are laborious, which is exactly the point: Section 7.3 will replace them with a two-line calculation.

Example 3.

Find $\int_{- 1}^{3} (2 - x) d x$ by the limit definition.

Solution

Let us divide the interval $[- 1, 3]$ into n subintervals of equal width $Δ x = \frac{3 - (- 1)}{n} = \frac{4}{n}$ . In each subinterval $[x_{k}, x_{k + 1}]$ , let us use $x_{k}^{*} = x_{k}$ as the sample point:

\begin{aligned} x_{0} &=-1\\ x_{1} &=-1+\Delta x=-1+\frac{4}{n}\\ x_{2} &=-1+2\Delta x=-1+2\left(\frac{4}{n}\right)\\ &\vdots\\ x_{k} &=-1+k\Delta x=-1+k\left(\frac{4}{n}\right)\\ &\vdots\\ x_{n} &=-1+n\Delta x=-1+n\left(\frac{4}{n}\right)=-1+4=3. \end{aligned}

Therefore $f (x_{k}^{*}) = f (x_{k}) = 2 - x_{k} = 2 - [- 1 + k (\frac{4}{n})] = 3 - \frac{4 k}{n}$ , and

\begin{aligned} \sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x &=\sum_{k=1}^{n}\left(3-\frac{4k}{n}\right)\frac{4}{n}\\ &=\sum_{k=1}^{n}\left(\frac{12}{n}-\frac{16k}{n^{2}}\right)\\ &=\frac{12}{n}\sum_{k=1}^{n}1-\frac{16}{n^{2}}\sum_{k=1}^{n}k\\ &=\frac{12}{n}(n)-\frac{16}{n^{2}}\cdot\frac{n(n+1)}{2}\\ &=12-8\left(1+\frac{1}{n}\right). \end{aligned}

Notice that in the above summation, n is a fixed number (say $n = 10$ ) and not a function of the summation index k. That is why we could move n outside the sigma sign. We have also used the formulas $\sum_{k = 1}^{n} 1 = n$ and $\sum_{k = 1}^{n} k = n (n + 1) / 2$ .

Because the width of each subinterval is the same, $max Δ_{k} x = 4 / n \to 0$ is equivalent to $n \to \infty$ , and we can conclude

\begin{aligned} \int_{-1}^{3}(2-x)\,dx &=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}^{*})\Delta x\\ &=\lim_{n\to\infty}\left[12-8\left(1+\frac{1}{n}\right)\right]\\ &=12-8(1+0)=4. \end{aligned}

You can easily show that if we use $x_{k}^{*} = x_{k + 1}$ instead of $x_{k}^{*} = x_{k}$ as the sample point in each subinterval $[x_{k}, x_{k + 1}]$ , we arrive at the same result.

Because the graph of $y = 2 - x$ is a straight line, we can check our answer with elementary geometry. We subtract the area of the part that lies below the $x$ -axis from the area of the part that is above the $x$ -axis. It follows from the triangle area formula $A = base \times height / 2$ that

\int_{- 1}^{3} (2 - x) d x = \frac{3 \times 3}{2} - \frac{1 \times 1}{2} = 4.

Graph of y equals 2 minus x showing a triangle above the x-axis and a smaller triangle below it — The integral of $2 - x$ from $- 1$ to $3$ is $\int_{- 1}^{3} (2 - x) d x = A_{1} - A_{2} = 9 / 2 - 1 / 2 = 4$ .

Example 4.

Find $\int_{1}^{3} x^{2} d x$ by the limit definition.

Solution

Let us divide the interval $[1, 3]$ into n subintervals of equal width $Δ_{k} x = Δ x = \frac{3 - 1}{n} = \frac{2}{n}$ :

\begin{aligned} x_{0} &=1\\ x_{1} &=1+\Delta x=1+\frac{2}{n}\\ x_{2} &=1+2\Delta x=1+2\left(\frac{2}{n}\right)\\ &\vdots\\ x_{k} &=1+k\Delta x=1+k\left(\frac{2}{n}\right)\\ &\vdots\\ x_{n} &=1+n\Delta x=1+n\left(\frac{2}{n}\right)=3. \end{aligned}

Take $f (x_{k}^{*}) = f (x_{k}) = {(1 + \frac{2 k}{n})}^{2} = 1 + \frac{4 k}{n} + \frac{4 k^{2}}{n^{2}}$ . Then

\begin{aligned} \sum_{k=1}^{n}f(x_{k}^{*})\Delta x &=\sum_{k=1}^{n}\left[\frac{2}{n}+\frac{8k}{n^{2}}+\frac{8k^{2}}{n^{3}}\right]\\ &=\frac{2}{n}\sum_{k=1}^{n}1+\frac{8}{n^{2}}\sum_{k=1}^{n}k+\frac{8}{n^{3}}\sum_{k=1}^{n}k^{2}\\ &=\frac{2}{n}\cdot n+\frac{8}{n^{2}}\cdot\frac{n(n+1)}{2}+\frac{8}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}\\ &=2+4\cdot\frac{n+1}{n}+\frac{4}{3}\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)\\ &=2+4\left(1+\frac{1}{n}\right)+\frac{4}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right). \end{aligned}

The width of every subinterval is the same, and $max Δ_{k} x \to 0$ is equivalent to $n \to \infty$ . As $n \to \infty$ , $\frac{1}{n} \to 0$ , and hence

\begin{aligned} \int_{1}^{3}x^{2}\,dx &=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}^{*})\Delta x\\ &=\lim_{n\to\infty}\left[2+4\left(1+\frac{1}{n}\right)+\frac{4}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right]\\ &=2+4(1+0)+\frac{4}{3}(1+0)(2+0)\\ &=\frac{26}{3}. \end{aligned}

There are no formulas from elementary geometry that can help us verify our answer, but we will soon learn a theorem, called the Fundamental Theorem of Calculus, that enables us to evaluate such definite integrals easily.

Region under the parabola y equals x squared between x equals 1 and x equals 3 — The region whose area is $\int_{1}^{3} x^{2} d x = 26 / 3$ .

Using Geometry Instead of the Limit

When the region under the graph is made of triangles, rectangles, or circular pieces, we can read off the integral without any limit at all.

Example 5.

Find $\int_{- 1}^{3} (| x | - 2) d x$ by interpreting the integral as the net area under the graph.

Solution

We know what the graph of $y = | x |$ looks like. To sketch the graph of $y = | x | - 2$ , we simply shift the graph of $y = | x |$ downward $2$ units. The net area of the region between the graph of $y = | x | - 2$ and the $x$ -axis over $[- 1, 3]$ can then be computed from triangles:

\begin{aligned} \int_{-1}^{3}(|x|-2)\,dx &=-A_{1}-A_{2}-A_{3}+A_{4}\\ &=-1-\frac{1}{2}-2+\frac{1}{2}\\ &=-3. \end{aligned}

The answer is negative because most of the region lies below the $x$ -axis.

Graph of y equals absolute value of x minus 2 with the regions above and below the x-axis labelled — The regions determining $\int_{- 1}^{3} (| x | - 2) d x$ .

Example 6.

Find $\int_{- 1}^{1} \sqrt{1 - x^{2}} d x$ using a geometric formula.

Solution

We recognize that the curve $y = \sqrt{1 - x^{2}}$ is the upper semicircle of radius 1 centered at the origin. Because we can easily compute the area between the semicircle and the $x$ -axis, taking the limit of Riemann sums is not required.

\begin{aligned} \int_{-1}^{1}\sqrt{1-x^{2}}\,dx &=\text{area between the semicircle of radius 1 and the }x\text{-axis}\\ &=\frac{1}{2}(\text{area of a circle of radius 1})\\ &=\frac{1}{2}\pi(1)^{2}=\frac{\pi}{2}. \end{aligned}

Upper semicircle of radius one centered at the origin, the graph of y equals the square root of 1 minus x squared — The graph of $y = \sqrt{1 - x^{2}}$ is a semicircle of radius 1.

Exercises

Exercise 1.

Let $f (x) = x^{2} + 1$ on the interval $[0, 2]$ . Using four subintervals of equal width, compute

(a) the left Riemann sum,

(b) the right Riemann sum,

(c) the midpoint Riemann sum.

Answer

(a) $\frac{15}{4} = 3.75$ (b) $\frac{23}{4} = 5.75$ (c) $\frac{37}{8} = 4.625$

Solution

Here $Δ x = \frac{2 - 0}{4} = \frac{1}{2}$ and the partition points are $0, 0.5, 1, 1.5, 2$ .

(a) Left endpoints are $0, 0.5, 1, 1.5$ :

\begin{aligned} S &= \tfrac{1}{2}\left[f(0)+f(0.5)+f(1)+f(1.5)\right]\\ &= \tfrac{1}{2}\left[1+1.25+2+3.25\right]=\tfrac{15}{4}. \end{aligned} TikZ figure

(b) Right endpoints are $0.5, 1, 1.5, 2$ :

\begin{aligned} S &= \tfrac{1}{2}\left[1.25+2+3.25+5\right]=\tfrac{23}{4}. \end{aligned} TikZ figure

(c) Midpoints are $0.25, 0.75, 1.25, 1.75$ :

\begin{aligned} S &= \tfrac{1}{2}\left[1.0625+1.5625+2.5625+4.0625\right]=\tfrac{37}{8}. \end{aligned} TikZ figure

The exact value is $\int_{0}^{2} (x^{2} + 1) d x = \frac{14}{3} \approx 4.667$ . As expected, the midpoint sum is by far the best of the three estimates, and because $f$ is increasing on $[0, 2]$ the left sum is the lower sum and the right sum is the upper sum. In the pictures this is visible at a glance: the left rectangles all sit below the curve, the right rectangles all poke above it, and the midpoint rectangles cut through it so that the overshoot and the undershoot nearly cancel.

Exercise 2.

For $f (x) = 4 - x^{2}$ on $[0, 2]$ with three subintervals of equal width, identify the upper Riemann sum and the lower Riemann sum, and compute both.

Answer

The upper sum is the left sum, $\frac{176}{27} \approx 6.52$ ; the lower sum is the right sum, $\frac{104}{27} \approx 3.85$ .

Solution

Here $Δ x = \frac{2}{3}$ , with subintervals $[0, \frac{2}{3}]$ , $[\frac{2}{3}, \frac{4}{3}]$ , $[\frac{4}{3}, 2]$ .

Since $f (x) = 4 - x^{2}$ is decreasing on $[0, 2]$ , the maximum on each subinterval occurs at its left endpoint and the minimum at its right endpoint. So the upper sum is the left sum, and the lower sum is the right sum. This reversal compared with the previous exercise is the whole point: "upper" does not mean "right."

Upper (left) sum:

\begin{aligned} U &= \tfrac{2}{3}\left[f(0)+f\!\left(\tfrac{2}{3}\right)+f\!\left(\tfrac{4}{3}\right)\right]\\ &= \tfrac{2}{3}\left[4+\tfrac{32}{9}+\tfrac{20}{9}\right] = \tfrac{2}{3}\cdot\tfrac{88}{9}=\tfrac{176}{27}\approx6.52. \end{aligned} TikZ figure

Lower (right) sum:

\begin{aligned} L &= \tfrac{2}{3}\left[f\!\left(\tfrac{2}{3}\right)+f\!\left(\tfrac{4}{3}\right)+f(2)\right]\\ &= \tfrac{2}{3}\left[\tfrac{32}{9}+\tfrac{20}{9}+0\right] = \tfrac{2}{3}\cdot\tfrac{52}{9}=\tfrac{104}{27}\approx3.85. \end{aligned} TikZ figure

The exact value $\int_{0}^{2} (4 - x^{2}) d x = \frac{16}{3} \approx 5.33$ lies between them, as it must. Compare the two pictures with those of the previous exercise: there the taller rectangles were on the right, here they are on the left, and the only thing that changed was whether $f$ increases or decreases.

Exercise 3.

Evaluate $\int_{0}^{2} (3 x + 1) d x$ directly from the limit definition, using right endpoints as the sample points.

Answer

$8$

Solution

With $Δ x = \frac{2}{n}$ and $x_{k} = \frac{2 k}{n}$ ,

\begin{aligned} \sum_{k=1}^{n}f(x_{k})\Delta x &= \sum_{k=1}^{n}\left(\frac{6k}{n}+1\right)\frac{2}{n}\\ &= \frac{12}{n^{2}}\sum_{k=1}^{n}k+\frac{2}{n}\sum_{k=1}^{n}1\\ &= \frac{12}{n^{2}}\cdot\frac{n(n+1)}{2}+\frac{2}{n}\cdot n\\ &= 6\left(1+\frac{1}{n}\right)+2. \end{aligned}

Letting $n \to \infty$ gives $6 + 2 = 8$ .

Check by geometry. The region is a trapezoid with parallel sides $f (0) = 1$ and $f (2) = 7$ and width 2, so its area is $\frac{1 + 7}{2} \times 2 = 8$ .

Exercise 4.

Evaluate $\int_{0}^{1} x^{2} d x$ from the definition, and state which summation formula you needed.

Answer

$\frac{1}{3}$ , using $\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ .

Solution

Take $Δ x = \frac{1}{n}$ and $x_{k} = \frac{k}{n}$ . Then

\begin{aligned} \sum_{k=1}^{n}x_{k}^{2}\,\Delta x &= \sum_{k=1}^{n}\frac{k^{2}}{n^{2}}\cdot\frac{1}{n} = \frac{1}{n^{3}}\sum_{k=1}^{n}k^{2}\\ &= \frac{1}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}\\ &= \frac{1}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right). \end{aligned}

As $n \to \infty$ this tends to $\frac{1}{6} (1) (2) = \frac{1}{3}$ .

Exercise 5.

Use geometry, not the limit definition, to evaluate each integral:

(a) $\int_{- 2}^{2} | x | d x$

(b) $\int_{0}^{3} \sqrt{9 - x^{2}} d x$

(c) $\int_{- 1}^{2} (1 - x) d x$

Answer

(a) $4$ (b) $\frac{9 π}{4}$ (c) $\frac{3}{2}$

Solution

(a) The graph consists of two triangles, each with base 2 and height 2, both above the axis: $2 \times \frac{2 \times 2}{2} = 4$ .

(b) $y = \sqrt{9 - x^{2}}$ is the upper semicircle of radius 3. From $x = 0$ to $x = 3$ we get one quarter of the disc: $\frac{1}{4} π (3)^{2} = \frac{9 π}{4}$ .

(c) The line $y = 1 - x$ crosses the axis at $x = 1$ . Above the axis on $[- 1, 1]$ we get a triangle of base 2 and height 2, area 2. Below the axis on $[1, 2]$ we get a triangle of base 1 and height 1, area $\frac{1}{2}$ , counted negatively: $2 - \frac{1}{2} = \frac{3}{2}$ .

Exercise 6.

Express the following limit as a definite integral:

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} \sqrt{1 + \frac{k}{n}} .

Answer

$\int_{0}^{1} \sqrt{1 + x} d x$ (equivalently $\int_{1}^{2} \sqrt{u} d u = \frac{2}{3} (2 \sqrt{2} - 1)$ ).

Solution

Read the sum against the template $\sum f (x_{k}) Δ x$ . Here $Δ x = \frac{1}{n}$ , which corresponds to an interval of total length 1, and the sample point is $x_{k} = \frac{k}{n}$ , a right endpoint on $[0, 1]$ . The remaining factor is $f (x_{k}) = \sqrt{1 + x_{k}}$ . Therefore

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} \sqrt{1 + \frac{k}{n}} = \int_{0}^{1} \sqrt{1 + x} d x .

The answer is not unique: with the substitution $u = 1 + x$ the same limit equals $\int_{1}^{2} \sqrt{u} d u$ . Any correct reading of the pattern is acceptable.

Exercise 7.

Suppose $\int_{0}^{4} f (x) d x = 7$ . Find

(a) $\int_{4}^{0} f (x) d x$

(b) $\int_{4}^{4} f (x) d x$

(c) $\int_{0}^{4} f (t) d t$

Answer

(a) $- 7$ (b) $0$ (c) $7$

Solution

(a) By the reversal rule, swapping the limits changes the sign.

(b) By the zero-width rule, an integral over an interval of length zero is zero.

(c) The variable of integration in a definite integral is a dummy variable, so renaming $x$ as $t$ changes nothing.

Exercise 8.

Sketch the graph of $f (x) = x - 1$ on $[0, 3]$ and use it to explain why $\int_{0}^{3} (x - 1) d x$ is positive even though part of the graph lies below the $x$ -axis. Then compute the value.

Answer

$\frac{3}{2}$ ; the area above the axis (2) exceeds the area below it $(\frac{1}{2})$ .

Solution

The line crosses the $x$ -axis at $x = 1$ . On $[0, 1]$ the graph lies below the axis, forming a triangle of base 1 and height 1, so its area is $\frac{1}{2}$ and it contributes $- \frac{1}{2}$ . On $[1, 3]$ the graph lies above the axis, forming a triangle of base 2 and height 2, area 2, contributing $+ 2$ . The net area is

\int_{0}^{3} (x - 1) d x = 2 - \frac{1}{2} = \frac{3}{2} > 0.

The integral measures net area, so a sign-changing integrand can still give a positive answer whenever the region above the axis is larger. The total geometric area, by contrast, would be $2 + \frac{1}{2} = \frac{5}{2}$ .

Exercise 9.

Explain why $\int_{0}^{1} f (x) d x$ need not exist if $f$ is unbounded on $[0, 1]$ , using $f (x) = 1 / x$ for $x > 0$ and $f (0) = 0$ as an illustration.

Answer

Because sample points can be chosen arbitrarily close to $0$ , the Riemann sums can be made arbitrarily large, so no single limiting value exists.

Solution

Fix any subdivision of $[0, 1]$ and consider the first subinterval $[0, x_{1}]$ . The sample point $x_{1}^{*}$ may be chosen anywhere in it, and $f (x_{1}^{*}) = 1 / x_{1}^{*}$ can be made as large as we please by taking $x_{1}^{*}$ close enough to zero. Its contribution to the Riemann sum is $f (x_{1}^{*}) x_{1}$ , which therefore has no upper bound, no matter how fine the subdivision is.

Since the sums do not approach a single number independent of the choice of the $x_{k}^{*}$ 's, the defining limit fails to exist and $f$ is not integrable on $[0, 1]$ in the sense of this section. (Integrals of this kind are treated separately as improper integrals.)

The moral: boundedness is necessary for Riemann integrability, and continuity on $[a, b]$ is a convenient sufficient condition.

Exercise 10.

Let $f$ be continuous and increasing on $[a, b]$ , and let $L_{n}$ and $R_{n}$ be the left and right Riemann sums with n equal subintervals. Show that

R_{n} - L_{n} = \frac{(b - a) [f (b) - f (a)]}{n},

and explain what this says about the accuracy of the estimates.

Answer

The difference telescopes to $Δ x [f (b) - f (a)]$ with $Δ x = \frac{b - a}{n}$ , so the gap between the two estimates shrinks like $1 / n$ .

Solution

With $Δ x = \frac{b - a}{n}$ and partition points $x_{0} = a, \dots, x_{n} = b$ ,

\begin{aligned} R_{n}-L_{n} &= \Delta x\sum_{k=1}^{n}f(x_{k})-\Delta x\sum_{k=1}^{n}f(x_{k-1})\\ &= \Delta x\sum_{k=1}^{n}\left[f(x_{k})-f(x_{k-1})\right]. \end{aligned}

The sum telescopes: every interior term $f (x_{k})$ appears once with a plus sign and once with a minus sign, leaving only $f (x_{n}) - f (x_{0}) = f (b) - f (a)$ . Hence

R_{n} - L_{n} = Δ x [f (b) - f (a)] = \frac{(b - a) [f (b) - f (a)]}{n} .

Because $f$ is increasing, $L_{n}$ is the lower sum and $R_{n}$ is the upper sum, so the true value of the integral is trapped between them. The width of that trap is proportional to $1 / n$ , which both proves that the two sums converge to a common limit and tells us that doubling the number of rectangles roughly halves the error.

Exercise 11.

Express the following limit as a definite integral, and evaluate it geometrically if you can:

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2}}{n^{3}} .

Answer

$\int_{0}^{1} x^{2} d x = \frac{1}{3}$

Solution

Force the expression into the shape $\sum f (x_{k}) Δ x$ by peeling off one factor of $\frac{1}{n}$ :

\frac{k^{2}}{n^{3}} = {(\frac{k}{n})}^{2} \cdot \frac{1}{n} .

So $Δ x = \frac{1}{n}$ , which signals an interval of total length 1, and the sample point is $x_{k} = \frac{k}{n}$ , the right endpoint of the $k$ -th subinterval of $[0, 1]$ . The remaining factor is $f (x_{k}) = x_{k}^{2}$ . Hence

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{k^{2}}{n^{3}} = \int_{0}^{1} x^{2} d x .

This is exactly the integral computed from the definition in Exercise 4, whose value is $\frac{1}{3}$ . Geometry alone will not give it, since the region under a parabola is not made of triangles and circular arcs, but the summation formula $\sum k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ does.

Exercise 12.

Express the following limit as a definite integral:

lim_{n \to \infty} \frac{π}{n} \sum_{k = 1}^{n} \sin \frac{k π}{n} .

What does the value of this integral say about the average height of one arch of the sine curve?

Answer

$\int_{0}^{π} \sin x d x = 2$ ; the average height of the arch is $\frac{2}{π} \approx 0.64$ .

Solution

The factor sitting outside the sine is $\frac{π}{n}$ , so $Δ x = \frac{π}{n}$ and the interval has total length $π$ . The sample point is

x_{k} = k Δ x = \frac{k π}{n},

which runs from $\frac{π}{n}$ up to $π$ : these are the right endpoints of the $n$ equal subintervals of $[0, π]$ . The remaining factor is $f (x_{k}) = \sin x_{k}$ . Therefore

lim_{n \to \infty} \frac{π}{n} \sum_{k = 1}^{n} \sin \frac{k π}{n} = \int_{0}^{π} \sin x d x = {[- \cos x]}_{0}^{π} = 2.

Notice the pattern in both this exercise and the previous one: the factor multiplying the sum is always $Δ x$ , and dividing the interval length by it recovers $n$ . Reading $Δ x$ first is the fastest way to identify the interval.

The average height. One arch of $y = \sin x$ sits over an interval of length $π$ and encloses area 2, so a rectangle of the same base with the same area would have height $\frac{2}{π} \approx 0.6366$ . That is the average value of $\sin x$ over $[0, π]$ , a quantity studied in Section 7.7. It is noticeably less than the peak height 1, because the arch is near its maximum only briefly.

Exercise 13.

Let $f (x) = x^{2} + 1$ on the interval $[- 1, 2]$ , divided into three subintervals of equal width. Compute the upper Riemann sum $U$ and the lower Riemann sum $L$ , and compare them with the left and right Riemann sums.

Answer

$U = 9$ , $L = 4$ , left sum $= 5$ , right sum $= 8$ . The exact value is $6$ . Neither the left nor the right sum is the upper or the lower sum.

Solution

Here $Δ x = \frac{2 - (- 1)}{3} = 1$ , and the subintervals are $[- 1, 0]$ , $[0, 1]$ , $[1, 2]$ .

The essential point is that $f$ is decreasing on $[- 1, 0]$ and increasing on $[0, 2]$ . So the location of the maximum and the minimum is not the same in every subinterval:

Subinterval	Behaviour of $f$	Maximum	Minimum
$[- 1, 0]$	decreasing	$f (- 1) = 2$ (left end)	$f (0) = 1$ (right end)
$[0, 1]$	increasing	$f (1) = 2$ (right end)	$f (0) = 1$ (left end)
$[1, 2]$	increasing	$f (2) = 5$ (right end)	$f (1) = 2$ (left end)

Upper sum. Take the maximum on each subinterval:

U = 1 \cdot [f (- 1) + f (1) + f (2)] = 2 + 2 + 5 = 9.

Lower sum. Take the minimum on each subinterval:

L = 1 \cdot [f (0) + f (0) + f (1)] = 1 + 1 + 2 = 4.

Comparison with the left and right sums.

left sum = f (- 1) + f (0) + f (1) = 2 + 1 + 2 = 5,

right sum = f (0) + f (1) + f (2) = 1 + 2 + 5 = 8.

The exact value is

\int_{- 1}^{2} (x^{2} + 1) d x = {[\frac{x^{3}}{3} + x]}_{- 1}^{2} = (\frac{8}{3} + 2) - (- \frac{1}{3} - 1) = 6.

All four estimates satisfy $L = 4 \leq 5, 6, 8 \leq 9 = U$ , as they must, but neither the left sum nor the right sum equals the upper or the lower sum. That coincidence happens only when $f$ is monotone on the whole interval, as in Exercises 1 and 2. Here the subinterval $[- 1, 0]$ takes its maximum at the left end while the other two take theirs at the right end, so no single endpoint rule can produce the upper sum.

Exercise 14.

Let $f (x) = \sin x$ on $[0, π]$ with four subintervals of equal width. Compute the upper and lower Riemann sums, and explain why the left and right Riemann sums are equal to each other yet equal to neither of them.

Answer

$U = \frac{π}{4} (2 + \sqrt{2}) \approx 2.681$ and $L = \frac{π \sqrt{2}}{4} \approx 1.111$ ; the left and right sums both equal $\frac{π}{4} (1 + \sqrt{2}) \approx 1.896$ . The exact value is $2$ .

Solution

Here $Δ x = \frac{π}{4}$ and the subintervals are $[0, \frac{π}{4}]$ , $[\frac{π}{4}, \frac{π}{2}]$ , $[\frac{π}{2}, \frac{3 π}{4}]$ , $[\frac{3 π}{4}, π]$ . The sine curve increases on $[0, \frac{π}{2}]$ and decreases on $[\frac{π}{2}, π]$ , so the two halves behave in opposite ways.

Subinterval	Behaviour	Maximum	Minimum
$[0, \frac{π}{4}]$	increasing	$\sin \frac{π}{4} = \frac{\sqrt{2}}{2}$	$\sin 0 = 0$
$[\frac{π}{4}, \frac{π}{2}]$	increasing	$\sin \frac{π}{2} = 1$	$\sin \frac{π}{4} = \frac{\sqrt{2}}{2}$
$[\frac{π}{2}, \frac{3 π}{4}]$	decreasing	$\sin \frac{π}{2} = 1$	$\sin \frac{3 π}{4} = \frac{\sqrt{2}}{2}$
$[\frac{3 π}{4}, π]$	decreasing	$\sin \frac{3 π}{4} = \frac{\sqrt{2}}{2}$	$\sin π = 0$

Upper sum.

\begin{aligned} U &= \frac{\pi}{4}\left[\frac{\sqrt{2}}{2}+1+1+\frac{\sqrt{2}}{2}\right]\\ &= \frac{\pi}{4}\left(2+\sqrt{2}\right)\approx2.6812. \end{aligned} TikZ figure

Lower sum.

\begin{aligned} L &= \frac{\pi}{4}\left[0+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+0\right]\\ &= \frac{\pi\sqrt{2}}{4}\approx1.1107. \end{aligned} TikZ figure

Notice that the first and last rectangles of the lower sum are degenerate: the minimum of $\sin x$ on $[0, \frac{π}{4}]$ and on $[\frac{3 π}{4}, π]$ is $0$ , so those rectangles collapse onto the $x$ -axis and contribute nothing. They are drawn as dashed segments above.

The left and right sums.

left sum = \frac{π}{4} [0 + \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2}] = \frac{π}{4} (1 + \sqrt{2}),

right sum = \frac{π}{4} [\frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 0] = \frac{π}{4} (1 + \sqrt{2}) .

They are equal, both approximately $1.896$ , because the list of sample values is the same list read backwards: the symmetry of the sine arch about $x = \frac{π}{2}$ makes $\sin$ take the same four values at the left endpoints as at the right endpoints.

Yet neither equals $U$ or $L$ . On the rising half the maximum sits at the right endpoint, while on the falling half it sits at the left endpoint, so the upper sum mixes the two rules; the lower sum mixes them the other way. The exact value is $\int_{0}^{π} \sin x d x = 2$ , and indeed

L \approx 1.111 \leq 1.896 \leq 2 \leq U \approx 2.681 .

The general lesson from this exercise and the previous one: upper and lower sums are defined by the behaviour of $f$ , not by a fixed choice of endpoint. They coincide with the right and left sums only when $f$ is monotone throughout $[a, b]$ .

Definition Of The Definite Integral

Riemann Sums

Some Specific Types of Riemann Sums

Limit of the Riemann Sums as $n \to \infty$

Dummy Variables

Reversing the Limits of Integration

Evaluating Integrals from the Definition

Using Geometry Instead of the Limit

Exercises

Quick Links

Social Media

Definition Of The Definite Integral

Riemann Sums

Some Specific Types of Riemann Sums

Limit of the Riemann Sums as n → ∞

Dummy Variables

Reversing the Limits of Integration

Evaluating Integrals from the Definition

Using Geometry Instead of the Limit

Exercises

Quick Links

Social Media

Limit of the Riemann Sums as $n \to \infty$