Definite Integral

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

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

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

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

Solution. \begin{align*}

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

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

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,
\int_0^1\sqrt{1+\cos x}dx&\leq\int_0^1\sqrt{2}dx\\
&\approx 1.4142136\\

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
Now we increase the number of sample points to infinity:
&=\frac{1}{b-a}\int_a^b f(x)dx.

Definition. The average or mean value of a continuous function $f(x)$ on $[a,b]$ is given by
\mathrm{av}(f)=\frac{1}{b-a}\int_a^b f(x)dx.

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,

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