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Definite Integrals

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 3.1: 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 $f(x)$ is called Riemann integrable function on $[a,b]$. We write $S$ as

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

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. $

Example 3..13   Find the are of rectangle bounded by the lines $y = x$, $y = 0$, and $x = 2$ using integration by quadrature.

SOLUTION Divide the interval $[0,2]$ into $2n$ 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,

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

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 $x = 3$

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

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

Riemann Integrable Functions

Theorem 3..7   If $f(x)$ is continuous on $[a,b]$, then $f(x)$ is a Riemann integrable function on $[a,b]$.

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 $c$ 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} $

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

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 $

3.
$\displaystyle \int_{a}^{b} f(x) dx$ $\displaystyle =$ $\displaystyle \lim_{\vert\Delta\vert \rightarrow 0}\sum_{i=1}^{n} f(\xi_{i})(x_{i} - x_{i-1})$  
  $\displaystyle =$ $\displaystyle \lim_{\vert\Delta\vert \rightarrow 0} - \sum_{i=1}^{n} f(\xi_{i})(x_{i-1} - x_{i}) = - \int_{b}^{a}f(x)\:dx \ensuremath{\ \blacksquare}$  

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 $x$.

Example 3..14   Let $f(t)$ be continuous. Then find

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

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 $

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 $u = 2x$. 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 $F$ is the indefinite integral of $f$ on $[a,b]$, 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} $

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.

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

2.

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

3.
$\displaystyle \int_0^2 \frac{dx}{x^2 + 4}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\tan^{-1}{\frac{x}{2}}\right]_0^2 = \frac{1}{2}\left(\tan^{-1}{\frac{2}{2}} - \tan^{-1}{0}\right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\cdot \frac{\pi}{4} = \frac{\pi}{8}\ensuremath{\ \blacksquare}$  

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

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 quadrature?Cevaluate $\displaystyle{\int_{0}^{1}x^{2}dx}$?D
2.
Evaluate the following integrals?D

(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?D

(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 $f(t)$ be continuous. Then find $g(x)$ ?D

(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?D

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

(c) $\displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{i}{n}}}$

Exercise B


1.
Suppose that $f(t)$ is continous function. Then find $g(x)$?D

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

(c) $\displaystyle{g(x) = \frac{d}{dx}\int_{0}^{2x}x^{2}f(t)dt}$

2.
Evaluate the following integrals?D

(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)?C(3)?C(4)?C(5)?D
4.
Prove the following inequality?D

(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?D

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

(c) $\displaystyle{\lim_{x \rightarrow 0} \int_{0}^{x} \tan{(t^2)}dt}$



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