# Integral Definition (The Definite Integral)

In this article, I discuss the definition of the definite integral and several properties of the definite integral. I cover many examples and also consider displacement and the area under a curve. So why is the definite integral so important?

We study the Riemann integral, also known as the Definite Integral. We consider its definition and several of its basic properties by working through several examples. The topics: displacement, the area under a curve, and the average value (mean value) are also investigated. We conclude with several exercises for more practice.

## The Definition of the Definite Integral

Suppose a bounded function $f$ is given along with a closed interval $[a,b]$ on which $f$ is defined. First we will partition the interval $[a,b]$ into $n$ subintervals by choosing points ${x_0, x_1, \ldots, x_n}$ arranged in such a way that \begin{equation} a=x_0 < x_1 < x_2 < \cdots <x_{n-1}< x_n =b. \end{equation} Call this partition $P.$ Now for $i=1, 2, \ldots, n$ the $i^{\text{th}}$ subinterval width is $\Delta_i=x_i-x_{i-1}.$ The largest of these widths is called the norm of the partition $P$ and is denoted by $\left| \left| P\right| \right|$; that is, $$\left| \left| P \right| \right|=\max_{i=1,2,\ldots,n} {\Delta_i}.$$ Arbitrarily choose a number from each subinterval, say $x_i^*.$ For $i=1,2,\ldots,n$ the number $x_i^*$ chosen from the $i^{\text{th}}$ is called the $i^{\text{th}}$ subinterval representative of the partition $P.$ Then the sum \begin{equation} \sum_{i=1}^n f(x_i^*) \Delta x_i \end{equation} is called the Riemann sum associated with $f$, the partition $P$, the chosen subinterval representatives.

Definition. Let $f$ be a bounded function defined on an interval $[a,b].$ Then the definite integral of $f$ over $[a,b]$ is defined as \begin{equation} \int_a^b f(x)\, dx := \lim_{\left|\left| P \right| \right| \to 0} \sum_{k=1}^n f(c_k) \Delta x. \end{equation} if the limit exists for all partitions $P$ of $[a,b]$ and all choices of $c_k$ in $[x_{k-1},x_k].$ When the above limit exists, we say the Riemann sums of $f$ on $[a,b]$ converge to the definite integral $I=\int_a^b f(x) \, dx$ and that $f$ is integrable over $[a,b].$

For special cases we can define particular kinds of Riemann sums. For example, if we choose equally spaced subintervals using $$\Delta x_i= \Delta x =\frac{b-a}{n}$$ and $x_i^*=a+k \Delta x$ for $i=1,2,\ldots,n$ then the Riemann sum formed is \begin{equation} \sum_{i=1}^n f(a+k \Delta x) \Delta x \end{equation} and the partition $P$ used to form this Riemann sum is called a regular partition. For regular partitions, $\left|\left| P\right|\right| \to 0$ if and only if $n\to \infty$ and so the definition of the definite integral sometimes is given as \begin{equation} \int_a^b f(x)\, dx := \lim_{n \to \infty } \sum_{k=1}^n f(c_k) \Delta x. \end{equation}

Theorem. If a function $f$ is continuous on an interval $[a,b]$ then its definite integral over $[a,b]$ exists.

Example. Express $$\lim_{\left| \left| |P \right| \right| \to 0 } \sum_{k=1}^n \sqrt{4-c_k^2} \Delta x_k$$ where $P$ is a partition of $[0,1]$, as a definite integral and find its value using geometry.

Solution. Recall the area of a circle with radius 2 is $4\pi.$ Let $f(x)=\sqrt{4-x^2}.$ Since $f$ is continuous on $[0,1]$, \begin{equation} \pi =\int_0^1 \sqrt{4-x^2}\, dx = \lim_{\left|\left|P\right|\right|\to 0 } \sum_{k=1}^n \sqrt{4-c_k^2} \Delta x_k. \end{equation} as desired.

Example. Express \begin{equation} \lim_{\left|\left|P\right|\right|\to 0 } \sum_{k=1}^n (\tan c_k) \Delta x_k \end{equation} where $P$ is a partition of $[0,\pi/4]$, as a definite integral and find its approximate value using geometry.

Solution. Let $f(x)=\tan x.$ Since $f$ is continuous on $[0,\pi/4]$, \begin{equation} \frac{\pi}{8}\approx \int_0^{\pi/4} \tan x\, dx = \lim_{\left|\left| P\right|\right| \to 0 } \sum_{k=1}^n (\tan c_k) \Delta x_k \end{equation} since the area under the curve is approximately equal to the area under the triangle with vertices $(0,0)$, $(\pi/4,0)$, and $(\pi/4,1).$

## Properties of the Definite Integral

Being familiar with the basic properties of finite sums and limits many of the properties listed below are straightforward to understand.

Theorem. Suppose $f$ and $g$ are integrable functions on $[a,b].$ Then

$(1) \quad \displaystyle \int_a^b f(x) \, dx =-\int_b^a f(x) \, dx$,

$(2) \quad \displaystyle \int_a^a f(x) \, dx =0$,

$(3) \quad \displaystyle \int_a^b k f(x) \, dx = k \int_a^b f(x) \, dx$,

$(4) \quad \displaystyle \int_a^b ( f(x) \pm g(x) ) \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$,

$(5) \quad \displaystyle \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$

(6) If $f$ has a maximum value $M$ and a minimum value $m$, then \begin{equation} m(b-a)\leq \int_a^b f(x) \, dx \leq M(b-a). \end{equation}

(7) If $f(x)\leq g(x)$ on $[a,b]$, then \begin{equation} \int_a^b f(x)\, dx \leq \int_a^b g(x) \, dx. \end{equation}

Example. Suppose $\displaystyle \int_0^2 f(x) \, dx=3$, $$\displaystyle \int_0^2 g(x) \, dx=-1 \quad \text{and}\quad \displaystyle \int_0^2 h(x) \, dx=3.$$ Evaluate \begin{equation} \int_0^2 [2f(x)+5g(x)-6h(x)] \, dx \end{equation} and then find a constant $s$ such that \begin{equation} \int_0^2 [5 f(x)+s g(x)-7 h(x)] \, dx=0. \end{equation}

Solution. Using the basic properties of the definite integral \begin{align} & \int_0^2 [2f(x)+5g(x)-7h(x)] \, dx \\ & \qquad =2 \int_0^2 f(x)\, dx +5\int_0^2g(x)\, dx -7\int_0^2h(x) \, dx \\ & \qquad = 2(3)+5(-1)-7(3)=-20. \end{align} Also
\begin{align} & \int_0^2 [5 f(x)+s \, g(x)-6 h(x)] \, dx \\ & \qquad = 5\int_0^2 f(x)\, dx +s \int_0^2 g(x)\, dx -7 \int_0^2h(x) \, dx \\ & \qquad =5(3) +s (-1) -6 (3) \, dx =-s-3=0 \end{align} which implies $s=-3.$

Example. Sketch the graph of $f$ and use it to evaluate $\displaystyle \int_{-1}^5 f(x) \, dx$ given \begin{equation} f(x)= \begin{cases} 2 & \text{for } -1\leq x \leq 1 \\ 3-x & \text{for } 1\leq x \leq 4 \\ 2x-9 & \text{for } 4\leq x \leq 5. \end{cases} \end{equation}

Solution. Using the basic properties of the definite integral \begin{align} \int_{-1}^5 f(x) \, dx & =\int_{-1}^1 f(x) \, dx +\int_{1}^4 f(x) \, dx+\int_{4}^5 f(x) \, dx \\ & =\int_{-1}^1 2 \, dx +\int_{1}^4 (3-x) \, dx+\int_{4}^5 (2x-9) \, dx \\ & = 2(2)+\frac{3}{2}-0=\frac{11}{2}. \end{align} as desired.

Example. Suppose that $f$ and $h$ are integrable, $\displaystyle \int_1^9 f(x) \, dx=-1$, $$\displaystyle \int_7^9 f(x) \, dx=5 \quad \text{and}\quad \displaystyle \int_7^9 h(x) \, dx=4.$$ Find the following values.

$(1) \quad \displaystyle \int_1^9 -2f(x) \, dx$

$(2) \quad \displaystyle \int_7^9 [f(x)+h(x)] \, dx$

$(3) \quad \displaystyle \int_7^9 [2f(x)-3h(x)] \, dx$

$(4) \quad \displaystyle \int_1^7 f(x) \, dx$

$(5) \quad \displaystyle \int_9^1 f(x) \, dx$

$(6) \quad \displaystyle \int_9^7 [h(x)-f(x)] \, dx$

Solution. Using the basic properties of the definite integral

$(1) \quad \displaystyle \int_1^9 -2f(x) \, dx =-2 \int_1^9 f(x) \, dx =-2 (-1)=2$

$(2) \quad \displaystyle \int_7^9 [f(x)+h(x)] \, dx =\int_7^9 f(x)\, dx +\int_7^9 h(x) \, dx =5+4=9$

$(3) \quad \displaystyle \int_7^9 [2f(x)-3h(x)] \, dx = 2 \int_7^9 f(x)\, dx -3\int_7^9h(x)] \, dx =-2$

$(4) \quad \displaystyle \int_1^7 f(x) \, dx =\int_1^9 f(x) \, dx -\int_7^9 f(x) \, dx =-1-5=-6$

$(5) \quad \displaystyle \int_9^1 f(x) \, dx =-\int_1^9 f(x) \, dx =-(-1)=1$

$(6) \quad \displaystyle \int_9^7 [h(x)-f(x)] \, dx =\int_7^9 f(x)\, dx-\int_7^9 h(x)] \, dx =5-4=1$

Example. What values for $a$ and $b$ minimize the following integral? $$\int_a^b(x^4-2x^2)\, dx$$

Solution. By solving $x^4-2x^2=0$ we determine that $x^4-2x^2\leq 0$ on $[-\sqrt{2},\sqrt{2}].$ Thus we find that $a=-\sqrt{2}$ and $b=\sqrt{2}$ will minimize the value of the given integral.

Example. (a) Find upper and lower bounds for $\displaystyle \int_0^1 \frac{1}{1+x^2} \, dx.$ (b) Find upper and lower bounds for $\displaystyle \int_0^{1/2} \frac{1}{1+x^2} \, dx$ and $\displaystyle \int_{1/2}^1 \frac{1}{1+x^2} \, dx.$ Use (b) to find another estimate of the integral in (a).

Solution. For (a) Since $f(x)=1/(1+x^2)$ is decreasing on $[0,1]$, it follows the maximum of $f$, $M$ occurs at 0 with $M=1$ and the minimum value of $f$, $m$ occurs at 1 with $m=1/2.$ Therefore, \begin{equation} \frac{1}{2}(1-0) =\frac{1}{2} \leq \int_0^1 \frac{1}{1+x^2} \, dx \leq 1 (1-0) =1. \end{equation} For (b) with $[0,1/2]$ we find, $M=1$ and $m=4/5$ and so \begin{equation}
\frac{4}{5} \left(\frac{1}{2}-0\right) =\frac{2}{5} \leq \int_0^{1/2} \frac{1}{1+x^2} \, dx \leq 1 (\frac{1}{2}-0) =\frac{1}{2}. \end{equation} For (b) with $[0,1/2]$ we find, $M=4/5$ and $m=1/2$ and so \begin{equation}
\frac{1}{2} \left(1-\frac{1}{2}\right) =\frac{1}{4} \leq \int_{1/2}^1 \frac{1}{1+x^2} \, dx \leq \frac{4}{5} (1-\frac{1}{2}) =\frac{2}{5} \end{equation} Then for part (c), we have a better estimate on the bounds of the original integral \begin{equation} \frac{13}{20}=\frac{1}{4}+\frac{2}{5}\leq \int_0^{1/2} \frac{1}{1+x^2} \, dx + \int_{1/2}^1 \frac{1}{1+x^2} \, dx \leq
\frac{1}{2}+\frac{2}{5}=\frac{9}{10}. \end{equation} namely, \begin{equation}
\frac{13}{20}\leq \int_0^1 \frac{1}{1+x^2} \, dx \leq \frac{9}{10}. \end{equation} as desired.

Example. Show that the value of $\displaystyle \int_0^1 \sqrt{x+8} \, dx$ lies between $2\sqrt{2}\approx 2.8$ and 3.

Solution. Since $f(x)=\sqrt{x+8}$ is increasing on $[0,1]$ it follows the maximum of $f$ is $f(1)=\sqrt{1+8}=3$ and the minimum of $f$ is $f(0) = \sqrt{0+8} = 2\sqrt{2}.$ Therefore, \begin{equation} 2\sqrt{2} =2\sqrt{2} (1-0)\leq \int_0^1 \sqrt{x+8} \, dx\leq 3 (1-0)=3. \end{equation} as desired.

## Displacement

Consider an object moving along a straight line. Assume the position of the object at time $t$ is given by the position function $s(t)$ and that its velocity at time $t$ is given by $v(t)=s'(t).$ If we happen to know that the object is always moving forward from $t=a$ to $t=b$, that is, $v(t)> 0$ on $[a,b]$, then the total distance travelled is $s(b)-s(a).$

Theorem. The total distance travelled by an object with continuous velocity $v(t)$ along a straight line from $t=a$ to $t=b$ is $$S=\int_a^b |v(t)| \, dt.$$

Example. Suppose $v(t)=1/(t+1)$ is the velocity of an object moving along a straight line. Use the formula $$S_n=\sum_{k=1}^n | v(a+k \Delta t) | \Delta t$$ where $\Delta t =(b-a)/n$ to estimate (using right endpoints) the total distance travelled by the object during the time interval $[0,1].$

Solution. Let $a=0$ with $\Delta t=1/4.$ Then \begin{equation}
v(a+k \Delta t)=v \left(\frac{k}{4}\right) = \frac{1}{1+\frac{k}{4}} =\frac{4}{k+4} \end{equation} Thus the total distance travelled by the object during the time interval $[0,1]$ is \begin{equation} S_4=\sum_{k=1}^4 \frac{4}{k+4}\left(\frac{1}{4}\right)\approx 0.635. \end{equation} as desired.

## Area Under a Curve

Theorem. Suppose $f$ is a continuous function and $f(x)\geq 0$ on the closed interval $[a,b].$ Then the area under the curve $y=f(x)$ on $[a,b]$ is given by the definite integral of $f$ on $[a,b].$

Example. Use the definition of the definite integral to find the area of the region between the curve $y=3x^2$ and the $x$-axis on the interval $[0,b].$

Solution. Let $\Delta x=\frac{b-0}{n}=\frac{b}{a}$ and let $x_0=0$, $x_1=\Delta x$, $x_2=2 \Delta x, \ldots, x_{n-1}=\Delta x$, $x_n=n \Delta x=b.$ Let the subinterval representatives, the $c_k’s$ be the right endpoints of the subintervals, so $c_1=x_1$, $c_2=x_2$, and so on. The rectangles defined have areas: \begin{align} & f(c_1)\Delta x = f(\Delta x)\Delta x =3(\Delta x)^2\Delta x=3 (\Delta x)^3 \\ & f(c_2)\Delta x = f(2\Delta x)\Delta x =3(2\Delta x)^2\Delta x=3 (2)^2 (\Delta x)^3 \\ & f(c_3)\Delta x = f(3\Delta x)\Delta x =3(3\Delta x)^2\Delta x=3 (3)^2 (\Delta x)^3 \\ & \vdots \\ & f(c_n)\Delta x = f(n\Delta x)\Delta x =3(n\Delta x)^2\Delta x=3 (n)^2 (\Delta x)^3 \end{align} Then forming the Riemann sum \begin{align} S_n & =\sum_{k=1}^n f(c_k) \Delta x = \sum_{k=1}^n 3 k^2(\Delta x) ^3 \\ & =3(\Delta x)^3 \sum_{k=1}^n k^2 =3\left(\frac{b^3}{n^3}\right) \left(\frac{n(n+1)(2n+1)}{6}\right) \\ & =\frac{b^3}{2}\left(2+\frac{3}{n}+\frac{1}{n^2}\right). \end{align} Therefore, applying the definition of the definite integral \begin{equation} \int_0^b 3x^2 \, dx =\lim_{n\to \infty } \frac{b^3}{2}\left(2+\frac{3}{n}+\frac{1}{n^2}\right) =b^3. \end{equation} as desired.

## Average Value

Let $f$ be a continuous function defined on $[a,b].$ We will ask the question: what are the average values that $f$ takes on over the interval $[a,b]$? To answer this, we will sample a finite number of values of $f.$ For example, say we sample $n$ values of $f$ by dividing $[a,b]$ into $n$ subintervals of equal width $\Delta x=\frac{b-a}{n}$ and evaluating $f$ at a point $c_k$ in each subinterval. The average of the $n$ sampled values is \begin{align} \frac{f(c_1)+f(c_2)+\cdots +f(c_n)}{n} & = \frac{1}{n} \sum_{k=1}^n f(c_k) \\ & = \frac{\Delta x}{b-a} \sum_{k=1}^n f(c_k) = \frac{1}{b-a} \sum_{k=1}^n f(c_k). \end{align}
If we pass the limit, if possible, as $n\to \infty$ we arrive at the following definition.

Definition. If $f$ is integrable on $[a,b]$ then its average value on $[a,b]$, denoted by the function $\text{av}(f)$, sometimes called its mean value, is \begin{equation} \text{av}(f) = \frac{1}{b-a}\int_a^b f(x) \, dx. \end{equation}

Example. Find the average value of the function $f(x)=3x^2-3$ over the interval $[0,1].$

Solution. The average value of the function $f(x)=3x^2-3$ over the interval $[0,1]$ is \begin{align} \left(\frac{1}{1-0}\right) \int_0^1 (3x^2-3)\, dx & =3 \int_0^1 x^2 \, dx – \int_0^1 3 \, dx \\ & =3\left(\frac{1}{3}\right)-3(1-0) =-2. \end{align} as desired.

Example. If $\text{av}(f)$ really is a typical value of the integrable function $f(x)$ on $[a,b]$, then the number $\text{av}(f)$ should have the same integral over $[a,b]$ that $f$ does. Show that $$\int_a^b \text{av}(f)\, dx= \int_a^b f(x) \, dx.$$

Solution. Notice that \begin{equation} \text{av}(f)=\frac{1}{b-a}\int_a^b f(x)\, dx:=K \end{equation} is a constant, say $K.$ Then \begin{align} \int_a^b K \, dx & = K \int_a^b \, dx =K (b-a) \\ & = (b-a) \frac{1}{b-a}\int_a^b f(x)\, dx =\int_a^b f(x) \, dx \end{align} and so yes the number $\text{av}(f)$ has the same integral over $[a,b]$ that $f$ does.

## Exercises on Integral Definition

Exercise. For each of the following you are given a function $f$ defined on an interval $[a,b]$, then number $n$ of subintervals of equal length ${\Delta x=(b-a)/n}$, and the evaluation points $c_k$ in $[x_{k-1}, x_k].$ For each of the following, (a) sketch the graph of $f$ and the rectangles associated with the Riemann sum for $f$ on $[a,b]$, and (b) find the Riemann sum.

$(1) \quad f(x)=2x-3, [0,2], n=4, c_k \text{ is the midpoint}$

$(2) \quad f(x)=-2x+1, [-1,2], n=6, c_k \text{ is the left endpoint}$

$(3) \quad f(x)=\sqrt{x}-1, [0,3], n=6, c_k \text{ is the right endpoint}$

$(4) \quad f(x)=2\sin x, [0,5\pi/4], n=5, c_k \text{ is the right endpoint}$

Exercise. Use the definition of the definite integral to evaluate the following integrals.

$(1) \quad \displaystyle \int_{-1}^2 x^2 \, dx$

$(2) \quad \displaystyle \int_{-1}^3 (x-2) \, dx$

$(3) \quad \displaystyle \int_{-1}^1 (2x+1) \, dx$

$(4) \quad \displaystyle \int_{-2}^1 (x^3+2x) \, dx$

Exercise. Each of the following is given as the limit of a Riemann sum of a function $f$ on $[a,b].$ Write this expression as a definite integral on $[a,b].$

$(1) \quad \displaystyle \lim_{n\to \infty} \sum_{k=1}^n (4 c_k-3)\Delta x, [-3,-1]$

$(2) \quad \displaystyle \lim_{n\to \infty} \sum_{k=1}^n 2c_k(1-c_k)^2\Delta x, [0,3]$

$(3) \quad \displaystyle \lim_{n\to \infty} \sum_{k=1}^n \frac{2c_k}{c_k^2+1} \Delta x, [1,2]$

$(4) \quad \displaystyle \lim_{n\to \infty} \sum_{k=1}^n c_k(\cos c_k)\Delta x, \left[0,\frac{\pi}{2}\right]$

Exercise. For each of the following make a sketch of $f$ on $[a,b]$ and then use the geometric interpretation of the integral to evaluate it.

$(1) \quad \displaystyle \int_{-2}^3 (2x+1) \, dx$

$(2) \quad \displaystyle \int_{-1}^2 |x| \, dx$

$(3) \quad \displaystyle \int_{-2}^2 \sqrt{4-x^2} \, dx$

$(4) \quad \displaystyle \int_{0}^2 \sqrt{-x^2+2x} \, dx$

Exercise. Given that $\int_{-1}^3 f(x) \, dx =4$ and $\int_3^6 f(x) \, dx=2$, evaluate the following integrals.

$(1) \quad \displaystyle \int_{-1}^3 [f(x)+g(x)] \, dx$

$(2) \quad \displaystyle \int_{-1}^3 [g(x)-f(x)] \, dx$

$(3) \quad \displaystyle \int_{-1}^3 [3f(x)-2g(x)] \, dx$

$(4) \quad \displaystyle \int_{3}^{-1} [f(x)+5g(x)] \, dx$

Exercise. Given that $\int_{-2}^2 f(x) \, dx =3$ and $\int_0^2 f(x) \, dx=2$, evaluate the following integrals.

$(1) \quad \displaystyle \int_{2}^0 f(x) \, dx$

$(2) \quad \displaystyle \int_{-2}^0 [f(x)+3] \, dx$

$(3) \quad \displaystyle \int_{2}^0 3f(x)\, dx-\int_0^{-2} 2f(x) \, dx$

Exercise. For each of the following use the properties of the integral to prove that the inequality without evaluating the integral.

$(1) \quad \displaystyle \int_0^1 \frac{\sqrt{x^3+x}}{x^2+1}\, dx \geq 0$

$(2) \quad \displaystyle \int_0^1 x^2\, dx \leq \int_0^1 \sqrt{x} \, dx$

$(3) \quad \displaystyle \int_0^{\pi/4} \sin^2 x \cos x \, dx \leq \int_0^{\pi/4} \sin^2 x \, dx$

$(4) \quad \displaystyle \int_0^{\pi/2} \cos x \, dx \leq \int_0^{\pi/2} (x^2+1) \, dx$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.