Definite Integrals

Understanding Partition the interval $[a,b]$ into smaller intervals. Consider the small rectangle with the base equal to the length of an interval and the height equal to the value of function.

Definite Integral

Suppose that $f(x)$ is defined on $[a,b]$ . Partition the interval $[a,b]$ into $n$ smaller intervals. Let $x_{i}$ be such that for $i = 1,2,\ldots,n$

**Figure 3.7:** Definite integral
$\includegraphics[width=7.1cm]{SOFTFIG-3/Fig3-7-1.eps}$

$\displaystyle a= x_{0} < x_{1}<x_{2}<\cdots< x_{i} < \cdots <x_{n} = b$

Express this partition by $\Delta$ . The norm of the interval is the width of intervals that is $\max_{0 \leq i \leq n} \Delta x_{i} = x_{i} - x_{i-1}$ and denoted by $\vert\Delta\vert$ . Now take $\xi_i$ in the subinterval $[x_{i-1},x_{i}]$ and consider the sum called Riemann Sum

$\displaystyle S(\Delta) = f(\xi_{1})\Delta x_{1} + f(\xi_{2})\Delta x_{2} + \cdots + f(\xi_{n})\Delta x_{n} = \sum_{i=1}^{n}f(\xi_{i})\Delta x_{i}$

$\displaystyle \lim_{\vert\Delta\vert \rightarrow 0}S(\Delta) = \lim_{\vert\Delta\vert \rightarrow 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} = S$

exists. Then we call $S$

definite integral of $f(x)$

and

is called Riemann integrable function on $[a,b]$

. We write $S$

$\displaystyle S = \lim_{\vert\Delta\vert \rightarrow 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} = \int_{a}^{b}f(x)dx$

Integrable Functions Note that $f(x)$ is integrable function on $[a,b]$ if and only if the limit of Riemann sum is independent of the choice of partition and choice of $\xi_{i}$ .

Integration by Quadrature Divide $[a,b]$ into $n(b-a)$ equal parts. Then $\Delta x = \frac{1}{n}$ and

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n(b-a)}f(\frac{i}{n}) = \int_{0}^{b-a}f(x) dx.$

Lower and Upper Limit To find the lower limit of an integral, Set $k = 1$ and $x = \lim_{n \to \infty}\frac{1}{n} = 0$ . To find the upper limit of an integral, Set $k = n(b-a)$ and $x = \lim_{n \to \infty}\frac{n(b-a)}{n} = b-a$ .

Example 3..13 Find the are of rectangle bounded by the lines $y = x$

, and

using integration by quadrature.

SOLUTION Divide the interval $[0,2]$

into

equal subintervals whose length is $\frac{1}{n}$ . Then Riemann sum is

$\displaystyle \lim_{n \to \infty}\big(\frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{2n}{n^2}\big).$

Rewrite using $\sum$ notation, we have

**Figure 3.8:** Example3-13
$\includegraphics[width=3cm]{SOFTFIG-3/quadrature1.eps}$

$\displaystyle \lim_{n \to \infty}\big(\frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{2n}{n^2}\big) = \lim_{n \to \infty}\sum_{k = 1}^{2n}\frac{k}{n^2}.$

Now take $\frac{1}{n}$ out of $\sum$ sign.

$\displaystyle \lim_{n \to \infty}\sum_{k = 1}^{2n}\frac{k}{n^2}$

$\displaystyle =$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{2n}\frac{k}{n}$

Let $x = \frac{k}{n}$ . Then $\begin{displaymath}\begin{array}{l\vert lll} k & 1 & \to & 2n\ \hline x & \frac{1}{n} & \to & 2 \end{array}\end{displaymath}$ Now let $n \to \infty$ . Then

$\displaystyle \lim_{n \to \infty}\sum_{k = 1}^{2n}\frac{k}{n^2}$

$\displaystyle =$

$\displaystyle \int_{x=0}^{2}x\;dx\ensuremath{ \blacksquare}$

Exercise 3..13 Express the area of the region bounded by $y = \frac{1}{x}$ , $y = 0$

, and the lines $x = 1$

and

SOLUTION Divide the interval $[1,3]$ into $2n$ equal subintervals whose length is $\frac{1}{n}$ . Then the Riemann sum is

**Figure 3.9:** Exercise3-13
$\includegraphics[width=3cm]{SOFTFIG-3/quadrature.eps}$

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\big(\frac{1}{1 + \frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \cdots + \frac{1}{1+\frac{2n}{n}}\big)$	$\displaystyle =$	$\displaystyle \frac{1}{n}\lim_{n \to \infty}\sum_{k=1}^{2n}\frac{1}{1 + \frac{k}{n}}$
	$\displaystyle =$	$\displaystyle \int_{x=0}^{2}\frac{1}{1+x}\;dx\ensuremath{ \blacksquare}$

Riemann Integrable Functions

Theorem 3..7 If is continuous on , then is a Riemann integrable function on .

Non Integrable Consider a function such that for $x$ a rational number, $f(x) = x$ and for $x$ an irrational number, $f(x) = 0$ . Then for every interval, there exists a rational number and an irrational number which implies that any Riemann sum can not converge

From the definition of definite integral, we can obtain the following formula.

Definite Integral Formula

Theorem 3..8 Let $f(x),g(x)$

be continuous on $[a,b]$

and

be a constant. Then

$\displaystyle{1. \int_{a}^{b}{f(x) \pm g(x)}dx = \int_{a}^{b}f(x)dx \pm \int_{a}^{b}g(x)dx }$

$\displaystyle{2. \int_{a}^{b}cf(x)dx = c\int_{a}^{b}f(x)dx }$

$\displaystyle{3. \int_{a}^{b}f(x) dx = - \int_{b}^{a} f(x) dx }$

$\displaystyle{4. \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx = \int_{a}^{b}f(x)dx}$

$\displaystyle{5. {\rm If} f(x) \geq g(x) {\rm for} [a,b], {\rm then} \int_{a}^{b}f(x)dx \geq \int_{a}^{b}g(x)dx }$

Rules 1. The integral of a sum is the sum of the integrals
2. The integral of a constant multiple is the constant multiple of the integral.
3. Interchange the lower limit and upper limit. Then put minus sign.
4.
5. Inequality is preserved after integration.

Proof 1. Suppose that $a < b$ . Then divide the interval $[a,b]$ by the partition

$\displaystyle a = x_{0} < x_{1} < \cdots < x_{n} = b$

Let $\xi_{i}$ be an element in $[x_{i-1},x_{i}]$ . Then consider the sum

$\displaystyle \sum_{i=1}^{n}\{f(\xi_{i}) + g(\xi_{i})\}\Delta x_{i} = \sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} + \sum_{i=1}^{n}g(\xi_{i})\Delta x_{i}$

Now $\vert\Delta\vert \rightarrow 0$ . Then by theorem 3.7,

$\displaystyle \int_{a}^{b}{f(x) + g(x)}dx = \int_{a}^{b}f(x)dx + \int_{a}^{b}g(x)dx$

$\displaystyle \int_{a}^{b} f(x) dx$	$\displaystyle =$	$\displaystyle \lim_{\vert\Delta\vert \rightarrow 0}\sum_{i=1}^{n} f(\xi_{i})(x_{i} - x_{i-1})$
	$\displaystyle =$	$\displaystyle \lim_{\vert\Delta\vert \rightarrow 0} - \sum_{i=1}^{n} f(\xi_{i})(x_{i-1} - x_{i}) = - \int_{b}^{a}f(x)\:dx \ensuremath{ \blacksquare}$

Let $f(x)$ be continuous on $[a,b]$ . Then consider the function expressed in the integral $\int_{a}^{x}f(t)dt$ , where $x$ is in $[a,b]$ . The domain of this function is $[a,b]$ .

2nd Fundamental Theorem of Integral Calculus

Theorem 3..9 Suppose that $f(x)$

is continuous on the closed interval $[a,b]$

. Then $\int_{a}^{x}f(t)dt$ is differentiable with respect to $x$

and

$\displaystyle \frac{d}{dx}\int_{a}^{x}f(t)dt = f(x) \hskip 1cm (a \leq x \leq b).$

NOTE Suppose that $f(t)$ is the speed of a particle. Then $\int_{a}^{x}f(t)dt$ can be thought of the distance traveled by the particle from the time $t = a$ to $t = x$ . $\frac{d}{dx}\int_{a}^{x}f(t)dt$ is the derivative at the time $x$ . In other wotds, the speed of a particle. Thus $f(t)$ is the speed of a particle at the time $t$ and $f(x)$ is the speed of a particle at the time $x$ .

Understanding
Let $F(x) = \int_{a}^{x}f(t)dt$ . Then , where is a primitive function of . Thus, a continuous function has a primitive function.

Example 3..14 Let $f(t)$

be continuous. Then find

$\displaystyle \frac{d}{dx}\int_{a}^{x^{2}}f(t)dt$

Note that $\frac{d}{dx}\int_a^x f(t)dt$ means $\frac{d(\int_a^x f(t)dt)}{dx}$ . SOLUTION Let $u = x^2$ . Then $du = 2x dx$ and

$\displaystyle \frac{d}{dx}\int_{a}^{x^{2}}f(t)dt = \frac{d (\int_{a}^{u} f(t) dt)}{du}\frac{du}{dx} = f(u)\frac{du}{dx} = f(x^2)(2x) \ensuremath{ \blacksquare}$

Exercise 3..14 Let $f(t)$

be continuous

$\displaystyle \frac{d}{dx}\int_{2x}^{2x+1}f(t)dt$

Limits of integration are from $2x$ to $2x+1$ and we can not apply the fundamental theorem of integration. Thus we divide the limits of integration. SOLUTION Let $c$ be a constant satisfying $2x < c < 2x+1$ . Then

$\displaystyle \int_{2x}^{2x+1}f(t)dt = \int_{2x}^{c}f(t)dt + \int_{c}^{2x+1}f(t)dt.$

Now $\int_{2x}^{c}f(t)dt = - \int_{c}^{2x}f(t)dt$

$\displaystyle \frac{d}{dx} \int_{2x}^{c}f(t)dt = -\frac{d}{dx} \int_{c}^{2x}f(t)dt.$

Let

. Then

$\displaystyle -\frac{d}{dx} \int_{c}^{2x}f(t)dt = -\frac{d(\int_{c}^{u}f(t)dt)}{du}\cdot \frac{du}{dx} = -f(u)\cdot 2 = -2f(2x).$

Next let $v = 2x+1$ . Then

$\displaystyle \frac{d}{dx}\int_{c}^{2x+1}f(t)dt = \frac{d(\int_c^v f(t)dt)}{dv}\cdot \frac{dv}{dx} = f(v) \cdot 2 = 2f(2x+1).$

Thus,

$\displaystyle \frac{d}{dx} \int_{2x}^{2x+1}f(t)dt = 2f(2x+1) - 2f(2x)\ensuremath{ \blacksquare}$

1st Fundamental Theorem of Integral Calculus

Theorem 3..10 Let $f(x)$

be a continuous function on the close interval $[a,b]$

and

is the indefinite integral of $f$

, then

$\displaystyle \int_{a}^{b}f(x)dx = G(b) - G(a)$

Proof By the 2nd fundamental theorem of integral calculus, $F(x) = \int_{a}^{x}f(t)dt$ is a primitive function of $f(x)$ . Thus, $F(x) = G(x) + c {\rm where} c {\rm is constant}$ . Since $F(a) = \int_a^a f(t)dt = 0$ , $F(a) = G(a) + c = 0$ . Thus, $c = -G(a)$ . From this, we have $F(x) = G(x) - G(a)$ . Set $x = b$ and we have

$\displaystyle F(b) = \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx = G(b) - G(a)\ensuremath{ \blacksquare}$

Notation We express $G(b) - G(a)$ as $[G(x)]_{a}^{b}$ .

Evaluation To evaluate a definite integral, we first find indefinite integral, then subtract the value obtained by substituting the lower limit from the value obtained by substituting the upper limit.

Example 3..15 Evaluate the following definite integrals.
1. $\displaystyle{\int_{1}^{4}\frac{1}{x^2}dx}$ 2. $\displaystyle{\int_{0}^{\frac{\pi}{2}}\sin^{2}{x}dx}$ 3. $\displaystyle{\int_0^2 \frac{dx}{x^2 + 4}}$

SOLUTION 1.

$\displaystyle \int_{1}^{4}\frac{1}{x^2}dx = \int_{1}^{4}x^{-2}dx = -\left[\frac... ...} \right]_{1}^{4} = -(\frac{1}{4} - 1) = \frac{3}{4}\ensuremath{ \blacksquare}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2}{x}dx$	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1-\cos{2x}}{2}dx = \frac{1}{2}\left[x - \frac{\sin{2x}}{2}\right]_{0}^{\frac{\pi}{2}}$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(\frac{\pi}{2}) = \frac{\pi}{4}\ensuremath{ \blacksquare}$

To evaluate $\left[x - \frac{\sin{2x}}{2}\right]_{0}^{\frac{\pi}{2}}$ , it is better to evaluate $\left[x \right]_0^{\frac{\pi}{2}} - \left[\frac{\sin{2x}}{2}\right]_{0}^{\frac{\pi}{2}}$ . This way, you have less mistakes.

$\displaystyle \int_0^2 \frac{dx}{x^2 + 4}$	$\displaystyle =$	$\displaystyle \frac{1}{2}\left[\tan^{-1}{\frac{x}{2}}\right]_0^2 = \frac{1}{2}\left(\tan^{-1}{\frac{2}{2}} - \tan^{-1}{0}\right)$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\cdot \frac{\pi}{4} = \frac{\pi}{8}\ensuremath{ \blacksquare}$

Note that if $\theta = \tan^{-1}{1}$ , then $tan{\theta} = 1$ , where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ . Thus $\theta = \frac{\pi}{4}$ .

Exercise 3..15 Evaluate the following definite integrals.
1. $\displaystyle{\int_{2}^{4} \frac{1}{x^2 + 2x - 3}\;dx}$ 2. $\displaystyle{\int_{1}^{2}\frac{x}{x^2 - 2x + 3}\;dx}$

SOLUTION 1. By partial fraction,

$\displaystyle \frac{1}{x^2 + 2x - 3} = \frac{1}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3}.$

Now clear the denominator,

$\displaystyle 1 = A(x+3) + B(x-1) = (A+B)x + 3A - B.$

From this, $A + B = 0, 3A - B = 1$

. Thus, $A = \frac{1}{4}, B = -\frac{1}{4}$ .

$\displaystyle \int_{2}^{4} \frac{1}{x^2 + 2x - 3}\;dx$	$\displaystyle =$	$\displaystyle \int_{2}^{4} \frac{1}{(x-1)(x+3)}\;dx$
	$\displaystyle =$	$\displaystyle \int_{2}^{4} \frac{1}{4}\big(\frac{1}{x-1} - \frac{1}{x+3}\big)\;dx$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\big[\log\vert x-1\vert -\log\vert{x+3}\vert\big]_2^4$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\big[\log\vert\frac{x-1}{x+3}\vert\big]_2^4$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\big(\log\frac{3}{7} - \log\frac{1}{5}\big) = \frac{1}{4}\log\frac{15}{7}\ensuremath{ \blacksquare}$

Check
$\log{\frac{3}{7}} - \log{\frac{1}{5}} = \log{\frac{\frac{3}{7}}{\frac{1}{5}}} = \log{\frac{3}{7}\cdot 5} = \log{\frac{15}{7}}$ .

Check
$\int \frac{g'(x)}{g(x)}\:dx = \log\vert g(x)\vert+c$ $\int \frac{1}{x^2 + a^2}\:dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})$ .

Check
Note $\tan^{-1}{\frac{1}{\sqrt{3}}} = \theta$ if and only if $\tan{\theta} = \frac{1}{\sqrt{3}} {\rm where} -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2})$ . Thus, $\theta = \frac{\pi}{6}$ .

2. The determinant $D$ of $x^2 - 2x + 4$ is given by $D = (-2)^2 - 4(4) = -12$ . Thus no more factorization.

$\displaystyle \int_{1}^{2}\frac{x}{x^2 - 2x + 4}\;dx$	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{1}^{2}\frac{2x-2+2}{x^2 - 2x + 4}\;dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_1^2 \frac{2x-2}{x^2 - 2x+4}\:dx + \frac{1}{2}\int_1^2 \frac{2}{x^2 - 2x+4}\:dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\big[\log\vert x^2 - 2x + 4\vert\big]_1^2 + \int_1^2 \frac{1}{x^2 -2x + 4}\:dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(\log{4} - \log{3}) + \int_1^2 \frac{1}{(x-1)^2 + 3}\:dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\log\frac{4}{3} + \left[\frac{1}{\sqrt{3}}\tan^{-1}{\frac{x-1}{\sqrt{3}}}\right]_1^2$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\log\frac{4}{3} + \frac{1}{\sqrt{3}}(\tan^{-1}(\frac{1}{\sqrt{3}}) - \tan^{-1}(0))$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\log\frac{4}{3} + \frac{\pi}{6\sqrt{3}} \ensuremath{ \blacksquare}$

Exercise A

1.

Using quadratureevaluate $\displaystyle{\int_{0}^{1}x^{2}dx}$

2.

Evaluate the following integrals

(a) $\displaystyle{\int_{0}^{1}(x^{2} + 3) dx}$ (b) $\displaystyle{\int_{1}^{2}\frac{x^{2} - 1}{x} dx}$ (c) $\displaystyle{\int_{0}^{1}\sqrt{x^{3}} dx}$ (d) $\displaystyle{\int_{0}^{\pi}\cos{x} dx}$ (e) $\displaystyle{\int_{0}^{\frac{\pi}{2}} \sin{x} dx}$

3.

Given $\displaystyle{\int_{0}^{1}f(x)dx = 6}$ , $\displaystyle{\int_{0}^{2}f(x)dx = 4}$ , and $\displaystyle{\int_{2}^{5}f(x)dx = 1}$ , find the following intgrals

(a) $\displaystyle{\int_{0}^{5}f(x) dx}$ (b) $\displaystyle{\int_{1}^{2}f(x) dx}$ (c) $\displaystyle{\int_{1}^{5}f(x) dx}$ (d) $\displaystyle{\int_{0}^{0}f(x) dx}$

(e) $\displaystyle{\int_{2}^{0}f(x) dx}$

4.

Let

be continuous. Then find $g(x)$

(a) $\displaystyle{g(x) = \frac{d}{dx}\int_{1}^{x}\sin{t}dt}$ (b) $\displaystyle{g(x) = \frac{d}{dx}\int_{x}^{1}\cos{t}dt}$ (c) $\displaystyle{g(x) = \frac{d}{dx}\int_{0}^{2x}\sqrt{\sin{t}}dt}$

5.

Evaluate the following limit

(a) $\displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{n} + \frac{2}{n} + \cdots + \frac{n}{n} \right)}$ (b) $\displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{2 + \frac{1}{n}} + \frac{1}{2 + \frac{2}{n}} + \cdots + \frac{1}{3} \right)}$

Exercise B

1.

Suppose that $f(t)$

is continous function. Then find $g(x)$

(a) $\displaystyle{g(x) = \frac{d}{dx}\int_{x}^{b}f(t)dt}$ (b) $\displaystyle{g(x) = \frac{d}{dx}\int_{x}^{x+1}f(t)dt}$

2.

Evaluate the following integrals

(a) $\displaystyle{\int_{1}^{5}2\sqrt{x-1}dx}$ (b) $\displaystyle{\int_{1}^{2}\frac{2-t}{t^{3}}dt}$ (c) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\cos{x}dx}$ (d) $\displaystyle{\int_{0}^{1}xe^{-x^{2}}dx}$

(e) $\displaystyle{\int_{0}^{\log{2}} \frac{e^{x}}{e^{x} + 1}dx}$

3.

Show the theorem 3.8(2)(3)(4)(5)

4.

Prove the following inequality

(a) $\displaystyle{\frac{\pi}{4} < \int_{0}^{1}\frac{1}{1 + x^n}dx < 1 (n > 2)}$ (b) $\displaystyle{\frac{1}{2n+2} \leq \int_{0}^{1} \frac{x^n}{1 + x}dx \leq \frac{1}{n} (n \geq 1)}$

5.

Find the limit of the followings

(a) $\displaystyle{\lim_{n \rightarrow \infty} \left(\frac{1}{n+1} + \frac{1}{n + 2} + \cdots + \frac{1}{2n} \right)}$ (b) $\displaystyle{\lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \sqrt{\frac{1}{n^2 + i^2}}}$