If $f: [a,b]\longrightarrow\mathbb{R}$ is a continuous function, the limits from the left-end point, midpoint, and right-end point methods in Areas under Curves exist and they are identical. The limit is denoted by $\int_a^b f(x)dx$ and called the *definite integral of $f(x)$ on the closed interval $[a,b]$*. It is not necessarily that $f(x)\geq 0$ for all $x\in [a,b]$, but if $f(x)\geq 0$ for all $x\in [a,b]$, then $\int_a^b f(x)dx$ is the area under the curve $y=f(x)$ on the interval $[a,b]$. It can be proven that the definite integral $\int_a^b f(x)dx$ does not depend on the choice of partition of $[a,b]$ or the choice of point in each subinterval to evaluate $f(x)$. More specifically, let $P: a=x_0<x_1<x_2<\cdots<x_n=b$ be an arbitrary partition (or a subdivision) of $[a,b]$ with $\Delta x_k=x_k-x_{k-1}$, $k=1,2,\cdots,n$. Also let $x_k’$ be any point in the subinterval $[x_{k-1},x_k]$, $k=1,2,\cdots,n$. Then

$$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^nf(x_k’)\Delta x_k.$$

The definite integral $\int_a^b f(x)dx$ satisfies the following properties.

*Theorem.* 1. $\int_a^b f(x)dx=-\int_b^a f(x)dx$.

2. $\int_a^a f(x)dx=0$.

3. $\int_a^b cf(x)dx=c\int_a^b f(x)dx$, where $c$ is a constant.

4. $\int_a^b (f(x)+g(x))dx=\int_a^b f(x)dx+\int_a^b g(x)dx$.

The properties 3 and 4 tell us that the definite integral $\int_a^b f(x)dx$ is linear.

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

6. Let $m$ and $M$ be the minimum and the Maximum values of $f(x)$ on $[a,b]$. Then

$$m(b-a)\leq\int_a^b f(x)dx\leq M(b-a).$$

7. If $f(x)\leq g(x)$ on $[a,b]$, then

$$\int_a^b f(x)dx\leq\int_a^b g(x)dx.$$

As a special case, if $f(x)\geq 0$, then $\int_a^b f(x)dx\geq 0$.

*Example*. Suppose that

$$\int_{-1}^1 f(x)dx=5,\ \int_1^4 f(x)dx=-2,\ \int_{-1}^1 h(x)dx=7.$$

Find

1. $\int_4^1 f(x)dx$.

*Solution*.

$$\int_4^1 f(x)dx=-\int_1^4 f(x)dx=2.$$

2. $\int_{-1}^1(2f(x)+3h(x))dx$.

*Solution*. \begin{align*}

\int_{-1}^1(2f(x)+3h(x))dx&=2\int_{-1}^1f(x)dx+3\int_{-1}^1h(x)dx\\

&=31.

\end{align*}

3. $\int_{-1}^4 f(x)dx$.

*Solution*.

\begin{align*}

\int_{-1}^4 f(x)dx&=\int_{-1}^1 f(x)dx+\int_1^4 f(x)dx\\

&=3.

\end{align*}

*Example*. Show that the value of $\int_0^1\sqrt{1+\cos x}dx$ is less than $\frac{3}{2}$.

*Solution*. Since $-1\leq\cos x\leq 1$ and $\sqrt{x}$ is an increasing function on $[0,\infty)$, $\sqrt{1+\cos x}\leq \sqrt{2}$. Hence,

\begin{align*}

\int_0^1\sqrt{1+\cos x}dx&\leq\int_0^1\sqrt{2}dx\\

&=\sqrt{2}\\

&\approx 1.4142136\\

&<\frac{3}{2}.

\end{align*}

Using the symmetries of even functions and odd functions, we obtain the following properties.

*Theorem*. Let $f:[-a,a]\longrightarrow\mathbb{R}$ be a continuous function. Then

1. If $f$ is an even function,

$$\int_{-a}^a f(x)dx=2\int_0^a f(x)dx.$$

2. If $f$ is an odd function,

$$\int_{-a}^a f(x)dx=0.$$

**An Application of Definite Integral: Average Value of a Continuous Functions**

Let $f: [a,b]\longrightarrow\mathbb{R}$ be a continuous function. Divide the closed interval $[a,b]$ into $n$ equal subintervals. Choose $n$ samples of $f(x)$ on $[a,b]$

$$f(c_1), f(c_2),\cdots,f(c_n)$$

such that $c_k\in[x_{k-1},x_k]$, $k=1,2,\cdots,n$. Then the average value of these $n$ samples is

\begin{align*}

\frac{f(c_1)+f(c_2)+\cdots+f(c_n)}{n}&=\frac{1}{b-a}(f(c_1)+f(c_2)+\cdots+f(c_n))\frac{(b-a)}{n}\\

&=\frac{1}{b-a}\sum_{k=1}^nf(c_k)\frac{(b-a)}{n}.

\end{align*}

Now we increase the number of sample points to infinity:

\begin{align*}

\lim_{n\to\infty}\frac{f(c_1)+f(c_2)+\cdots+f(c_n)}{n}&=\lim_{n\to\infty}\frac{1}{b-a}\sum_{k=1}^nf(c_k)\frac{(b-a)}{n}\\

&=\frac{1}{b-a}\int_a^b f(x)dx.

\end{align*}

*Definition*. The average or mean value of a continuous function $f(x)$ on $[a,b]$ is given by

\begin{equation}

\label{eq:mv}

\mathrm{av}(f)=\frac{1}{b-a}\int_a^b f(x)dx.

\end{equation}

*Example*. Find the average value of $f(x)=\sqrt{4-x^2}$ on $[-2,2]$.

Solution. Notice that $y=\sqrt{4-x^2}$ on $[-2,2]$ is the upper semi-circle centered at the origin with radius 2. Hence,

\begin{align*}

\mathrm{av}(f)&=\frac{1}{2-(-2)}\int_{-2}^2\sqrt{4-x^2}dx\\

&=\pi.

\end{align*}