maths > integral-calculus

Integration: First Principles

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Overview

Integration: First Principles :

»  cause-effect relation in two quantities

→  eg: speed-displacement

»  The cause is calculated as a function of an algebraic expression in a variable.

→  eg: speed $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$

»  The effect is derived to be "continuous aggregate of cause with respect to the variable".

→  eg : displacement = continuous aggregate of displacement

»  In such a case, the effect is another algebraic expression in the variable.

→  displacement is a function of $t$$t$

»  The effect is computed as continuous aggregate : the sum of change over an interval of the variable.

→  displacement $=\underset{n\to \infty }{lim}\sum _{i=1}^{n}v\left(\frac{ix}{n}\right)\frac{x}{n}$$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} v \left(\frac{i x}{n}\right) \frac{x}{n}$

»  Integration or integral of a function

In an interval from $0$$0$ to $x$$x$, the continuous aggregate of function $f\left(x\right)$$f \left(x\right)$ is the integral of the function.

$\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$

$=c+{\int }_{0}^{x}f\left(x\right)dx$$= c + {\int}_{0}^{x} f \left(x\right) \mathrm{dx}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)\frac{x}{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \frac{x}{n}$

to aggregate means to combine

A car travels at speed $10m/s$$10 m / s$. The distance traveled in $2$$2$ seconds is "$20m$$20 m$". Speed multiplied by time gives the distance traveled in that time.

A car travels at speed $10m/s$$10 m / s$ in the first $2$$2$ seconds, and at speed $15m/s$$15 m / s$ in the next $3$$3$ seconds. The total distance traveled is " $=10×2+15×3=65$$= 10 \times 2 + 15 \times 3 = 65$ m"

Distance is computed as speed (cause) repeatedly added for the time durations (aggregate).

distance = speed1 $×$$\times$ time1 $+$$+$ speed2 $×$$\times$ time2 $+...$$+ \ldots$

Distance is the aggregate of speed. The term aggregate is noted.

function of a variable

A car is moving at speed $20$$20$m/sec. What is the distance covered in $t$$t$ sec ? can it be given without a numerical value for $t$$t$. Yes, "For any value of $t$$t$, the distance traveled is $s=20t$$s = 20 t$ meter".

Measurement can be expressed as a function of a variable.

substitute

A car is moving with speed given as a function of time $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$ m/sec. The speed at $t=2\mathrm{sec}$$t = 2 \sec$ is "$14$$14$ meter/sec". To get substitute $t=2$$t = 2$ in the formula for $v$$v$.

varying with time

A car is moving with speed given as a function of time $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$ m/sec. Couple of students want to calculate the distance traveled in $2$$2$ sec. We know that

$\text{distance}=\text{speed}×\text{time}$$\textrm{\mathrm{di} s \tan c e} = \textrm{s p e e d} \times \textrm{t i m e}$

•  Person A does the following: At $t=2$$t = 2$s, the speed is $14$$14$m/sec. So the distance is $14×2=28$$14 \times 2 = 28$m.

•  Person B does the following: At $t=1$$t = 1$s, the speed is $5$$5$m/sec. And at $t=2$$t = 2$s, the speed is $14$$14$m/sec. So the distance traveled is $5×1+14×\left(2-1\right)=19$$5 \times 1 + 14 \times \left(2 - 1\right) = 19$ m.

Which one is correct? Neither of them is.

It is very important to understand the above problem formulation and the time-interval solution here. The speed varies with time, and the distance travelled also varies with time. The best one can do (with the given tool $\text{distance}=\text{speed}×\text{time}$$\textrm{\mathrm{di} s \tan c e} = \textrm{s p e e d} \times \textrm{t i m e}$ ) is to go for small time intervals and find approximate speed.

time-intervals approximation

$\text{distance}=\text{speed}×\text{time}$$\textrm{\mathrm{di} s \tan c e} = \textrm{s p e e d} \times \textrm{t i m e}$

One student wants to calculate the distance traveled.

•  At $t=0$$t = 0$ second, the speed is $=3×0+2=2$$= 3 \times 0 + 2 = 2$m/sec.

•  At $t=1$$t = 1$ second, the speed is $=3×1+2=5$$= 3 \times 1 + 2 = 5$m/sec.

•  At $t=2$$t = 2$ second, the speed is $=3×4+2=14$$= 3 \times 4 + 2 = 14$m/sec.

Speed varies with time. So, consider time intervals $0$$0$ to $1$$1$sec and $1$$1$sec to $2$$2$sec. In each of these time intervals, the average of start and end speed can be taken as the speed during the time interval.

The speed during time $0$0 < t < 1$ $=\frac{2+5}{2}=3.5$$= \frac{2 + 5}{2} = 3.5$ m/sec

The speed during time $1$1 < t < 2$ $=\frac{5+14}{2}=9.5$$= \frac{5 + 14}{2} = 9.5$ m/sec

So distance traveled is $=3.5×1+9.5×\left(2-1\right)=13m$$= 3.5 \times 1 + 9.5 \times \left(2 - 1\right) = 13 m$

With this "Approximate value of distance traveled" is computed. It is approximate, because the speed continuously changes with time, but the calculation approximates to average over $1$$1$ second intervals.

Note that the above is a form of aggregate.

derive to an expression

But, A car is moving with speed given as a function of time $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$ or an algebraic expression. In that case, we expect that "displacement is a function of time" or a derived algebraic expression.

In this, The displacement is aggregate of speed. So, one has to figure out how to find the aggregate of an algebraic expression into a derived expression.

formulation

A car is moving with speed given as a function of time $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$. Displacement is aggregate of speed over time intervals. Few students are set to find the aggregate as a function of time.
students may work these out to understand

•  Person A found continuous aggregate, displacement, for $1$$1$ second interval. $\sum _{i=1,2,3,...}^{t}\left(3{i}^{2}+2\right)×1$${\sum}_{i = 1 , 2 , 3 , \ldots}^{t} \left(3 {i}^{2} + 2\right) \times 1$

•  Person B found continuous aggregate, displacement, for $2$$2$ second interval. $\sum _{i=2,4,6,...}^{t}\left(3{i}^{2}+2\right)×2$${\sum}_{i = 2 , 4 , 6 , \ldots}^{t} \left(3 {i}^{2} + 2\right) \times 2$

•  Person C found continuous aggregate, displacement, for $0.5$$0.5$ second interval. $\sum _{i=0.5,1,1.5,...}^{t}\left(3{i}^{2}+2\right)×0.5$${\sum}_{i = 0.5 , 1 , 1.5 , \ldots}^{t} \left(3 {i}^{2} + 2\right) \times 0.5$

The above calculates approximation of the distance traveled. Note that the summation goes uptill time $t$$t$ and so the expression is a function of $t$$t$ or in other words, an algebraic expression

error in formulation

The approximate distance can be computed with time interval $\delta$$\delta$ is
$\sum _{i=\delta ,2\delta ,3\delta ,...}^{t}\left(3{i}^{2}+2\right)×\delta$${\sum}_{i = \delta , 2 \delta , 3 \delta , \ldots}^{t} \left(3 {i}^{2} + 2\right) \times \delta$
where $\delta =\frac{t}{n}$$\delta = \frac{t}{n}$,
$n$$n$ is the number of steps between $0$$0$ and $t$$t$

The error in the approximation is reduced when the number of steps $n$$n$ is large. The speed changes with time. As the step-size is made smaller, the speed is approximated better. For smaller step size, the total number of steps has to be higher.

continous aggregate

A car is moving with speed given as a function of time $v=3{t}^{2}+2$$v = 3 {t}^{2} + 2$. An odometer is attached to a wheel. The odometer measures the rotation of the wheel and proportionally provides the distance traveled by the car. The distance shown in the odometer is "continuous-aggregate distance traveled for any given time".

from apprx to accurate

The approximate distance can be computed with time interval $\delta$$\delta$ is
$\sum _{i=\delta ,2\delta ,3\delta ,...}^{t}\left(3{i}^{2}+2\right)×\delta$${\sum}_{i = \delta , 2 \delta , 3 \delta , \ldots}^{t} \left(3 {i}^{2} + 2\right) \times \delta$
where $\delta =\frac{t}{n}$$\delta = \frac{t}{n}$,
$n$$n$ is the number of steps between $0$$0$ and $t$$t$

The algebraic expression exactly the continuous-aggregate distance when the number of steps $n$$n$ is $\infty$$\infty$ or when the step width $\delta$$\delta$ is $0$$0$

This is a big jump in understanding. The speed is continuously changing and the summation is done continuously (not in steps) when $n$$n$ is $\infty$$\infty$.

generalizing

Generalizing that, for a function $f\left(x\right)$$f \left(x\right)$ the approximate continuous aggregate is $\sum _{i=\delta ,2\delta ,3\delta ,...}^{x}f\left(i\right)×\delta$${\sum}_{i = \delta , 2 \delta , 3 \delta , \ldots}^{x} f \left(i\right) \times \delta$
where $\delta =\frac{x}{n}$$\delta = \frac{x}{n}$

The same can be given as $\sum _{i=1,2,3,...}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$${\sum}_{i = 1 , 2 , 3 , \ldots}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$
where $i$$i$ takes positive integer values.

The same is simplified as $\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$${\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

direct substitution

When $n=\infty$$n = \infty$, the expression gives continuous-aggregate distance.
$\delta =\frac{t}{n}=0$$\delta = \frac{t}{n} = 0$

$\text{distance}$$\textrm{\mathrm{di} s \tan c e}$$=\sum _{i=1}^{\infty }\left(3×{0}^{2}+2\right)×0$$= {\sum}_{i = 1}^{\infty} \left(3 \times {0}^{2} + 2\right) \times 0$
$=0+0+0+...\infty \phantom{\rule{1ex}{0ex}}\text{times}$$= 0 + 0 + 0 + \ldots \infty \textrm{\times}$
$=0×\infty$$= 0 \times \infty$

The value of $0×\infty$$0 \times \infty$ is "indeterminate value $0/0$$0 / 0$".

How does one solve a function evaluating to indeterminate value?
The answer is "Use Limit of the function as $n$$n$ approaching $\infty$$\infty$."

limit

A car is moving with speed given as a function of time $v\left(t\right)=3{t}^{2}+2$$v \left(t\right) = 3 {t}^{2} + 2$. The approximate distance $=\sum _{i=1}^{n}\left(3{\left(\frac{it}{n}\right)}^{2}+2\right)×\frac{t}{n}$$= {\sum}_{i = 1}^{n} \left(3 {\left(\frac{i t}{n}\right)}^{2} + 2\right) \times \frac{t}{n}$
$n$$n$ is the number of steps between $0$$0$ and $t$$t$

When $n=\infty$$n = \infty$, the expression gives continuous-aggregate distance
$=\sum _{i=1}^{n}\left(3{\left(\frac{it}{n}\right)}^{2}+2\right)×\frac{t}{n}{\mid }_{n=\infty }$$= {\sum}_{i = 1}^{n} \left(3 {\left(\frac{i t}{n}\right)}^{2} + 2\right) \times \frac{t}{n} {|}_{n = \infty}$
$=\sum _{i=1,2...}^{n}0$$= {\sum}_{i = 1 , 2. . .}^{n} 0$
$=0×\infty =\frac{0}{0}$$= 0 \times \infty = \frac{0}{0}$

Since the distance evaluates to indeterminate value, the limit is used to check if the function is defined.

$\underset{n\to \infty }{lim}\sum _{i=1}^{n}$${\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(3{\left(\frac{it}{n}\right)}^{2}+2\right)×\frac{t}{n}$$\left(3 {\left(\frac{i t}{n}\right)}^{2} + 2\right) \times \frac{t}{n}$

$=\underset{n\to \infty }{lim}\sum _{i=1}^{n}$$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(3{\left(\frac{it}{n}\right)}^{2}\right)×\frac{t}{n}$$\left(3 {\left(\frac{i t}{n}\right)}^{2}\right) \times \frac{t}{n}$$+\underset{n\to \infty }{lim}\sum _{i=1}^{n}\left(2\right)×\frac{t}{n}$$+ {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(2\right) \times \frac{t}{n}$

$=\underset{n\to \infty }{lim}\frac{3{t}^{3}}{{n}^{3}}\sum _{i=1}^{n}{i}^{2}$$= {\lim}_{n \to \infty} \frac{3 {t}^{3}}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2}$$+\underset{n\to \infty }{lim}\frac{2t}{n}\sum _{i=1}^{n}1$$+ {\lim}_{n \to \infty} \frac{2 t}{n} {\sum}_{i = 1}^{n} 1$

substituting $\sum _{i=1}^{n}{i}^{2}=n\left(2n+1\right)\left(n+1\right)/6$${\sum}_{i = 1}^{n} {i}^{2} = n \left(2 n + 1\right) \left(n + 1\right) / 6$
and $\sum _{i=1}^{n}1=n$${\sum}_{i = 1}^{n} 1 = n$
$=\underset{n\to \infty }{lim}\frac{3{t}^{3}}{{n}^{3}}×\frac{n\left(2n+1\right)\left(n+1\right)}{6}$$= {\lim}_{n \to \infty} \frac{3 {t}^{3}}{n} ^ 3 \times \frac{n \left(2 n + 1\right) \left(n + 1\right)}{6}$$+\underset{n\to \infty }{lim}\frac{2t}{n}×n$$+ {\lim}_{n \to \infty} \frac{2 t}{n} \times n$

$=\underset{n\to \infty }{lim}\frac{3{t}^{3}}{{n}^{3}}×\frac{2{n}^{3}+3{n}^{2}+n}{6}$$= {\lim}_{n \to \infty} \frac{3 {t}^{3}}{n} ^ 3 \times \frac{2 {n}^{3} + 3 {n}^{2} + n}{6}$$+\underset{n\to \infty }{lim}\left(2t\right)$$+ {\lim}_{n \to \infty} \left(2 t\right)$

$=\underset{n\to \infty }{lim}\frac{3{t}^{3}}{6}×\left(2+\frac{3}{n}+\frac{1}{{n}^{2}}\right)$$= {\lim}_{n \to \infty} \frac{3 {t}^{3}}{6} \times \left(2 + \frac{3}{n} + \frac{1}{n} ^ 2\right)$$+\underset{n\to \infty }{lim}2t$$+ {\lim}_{n \to \infty} 2 t$

applying limit
$=\frac{3{t}^{3}}{6}×\left(2+0+0\right)+2t$$= \frac{3 {t}^{3}}{6} \times \left(2 + 0 + 0\right) + 2 t$

$={t}^{3}+2t$$= {t}^{3} + 2 t$

The continuous-aggregate distance is computed as an algebraic expression.

initial value

Note that the distance traveled is computed starting from time $t=0$$t = 0$, at which point, the initial distance of the car can be non-zero. So,
the distance = initial distance at time $0$$0$ + distance traveled between time $0$$0$ and $t$$t$.
For the given problem, the distance $=c$$= c$$+\underset{n\to \infty }{lim}\sum _{i=1}^{n}$$+ {\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(3{\left(\frac{it}{n}\right)}^{2}+2\right)×\frac{t}{n}$$\left(3 {\left(\frac{i t}{n}\right)}^{2} + 2\right) \times \frac{t}{n}$
where $c$$c$ is a constant.

summary

Summarizing the learning so far,

•  Two quantities are in a cause-effect relation.

•  the cause is calculated as a function of an algebraic expression in a variable.

•  The effect is derived to be "continuous aggregate of cause with respect to the variable".
(note: there are other forms of relation between cause-effect, such as multiple, addition, exponent. In this topic, we are concerned with only the continuous aggregate relation.)

•  In such a case, the effect is another algebraic expression in the variable.

•  The effect is computed as, the continuous aggregate: the aggregate of the cause, over an interval, with the interval split into infinite partitions. This calculation is named as integration or integral of the function.

finer details

Note 1: The summation in integral has many other forms, Riemann, Lebesgue, and Darboux forms, which will be introduced in due course.

Note 2: Differentiation is instantaneous rate of change or rate-of-change. Integration is continuous aggregate or aggregate

The word 'integrate' means: combine one with another to form a whole.

The continuous aggregate is aptly named integral or integration.

Integration in the context of cause-effect relation in continuous aggregate relation:
If the cause is given by $f\left(x\right)$$f \left(x\right)$ then the effect is computed as integration or integral of $f\left(x\right)$$f \left(x\right)$ denoted as
$\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

The second term is understood to be continuous aggregate between $0$$0$ and $x$$x$. This is denoted as
$\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$

$=c+{\int }_{0}^{x}f\left(x\right)dx$$= c + {\int}_{0}^{x} f \left(x\right) \mathrm{dx}$
$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

definition

Cause-effect was explained to understand the physical significance. Abstracting this and understanding the quantities involved in integration:

A quantity $u=f\left(x\right)$$u = f \left(x\right)$ is related to another quantity $v$$v$ such that $v$$v$ is the continuous aggregate of $u$$u$ with respect to $x$$x$. Then,
$v$$v$

$=\int udx$$= \int u \mathrm{dx}$

$=c+{\int }_{0}^{x}udx$$= c + {\int}_{0}^{x} u \mathrm{dx}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

Note that $\int udx$$\int u \mathrm{dx}$ is another quantity $v$$v$, related to the given quantity $u$$u$.

• $f\left(x\right)$$f \left(x\right)$ is called integrand.

• $x$$x$ is the variable of integration.

• $c$$c$ is the constant of integration.

• in ${\int }_{0}^{x}$${\int}_{0}^{x}$, the value $0$$0$ denotes the start position of integration called "lower limit" of integration.

• in ${\int }_{0}^{x}$${\int}_{0}^{x}$, the variable $x$$x$ denotes the end position of integration called "upper limit" of integration.

understanding limit

Integration is the continuous-aggregate given by $c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$$c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

• $\underset{n\to \infty }{lim}$${\lim}_{n \to \infty}$ represents "continuous" (as against the discrete number of steps between the start and end positions of summation).

• $\sum _{i=1}^{n}$${\sum}_{i = 1}^{n}$ represents "summation" or aggregate;

• $\frac{x}{n}$$\frac{x}{n}$ represents the $n$$n$ partitions between the start and end positions.

• $f\left(\frac{ix}{n}\right)$$f \left(\frac{i x}{n}\right)$ represents the value of function in the ${i}^{\text{th}}$${i}^{\textrm{t h}}$ partition.

• $f\left(\frac{ix}{n}\right)×\frac{x}{n}$$f \left(\frac{i x}{n}\right) \times \frac{x}{n}$ represents the value contributed by the ${i}^{\text{th}}$${i}^{\textrm{t h}}$ partition

understanding integral

Students can connect the notation $\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$ as

•  the small difference in $x$$x$ is given as $dx$$\mathrm{dx}$

•  the multiplication by $f\left(x\right)$$f \left(x\right)$ to $dx$$\mathrm{dx}$ implies, the value of function for the small difference is multiplied.

•  $\int$$\int$ denotes the sum with limit of $dx$$\mathrm{dx}$ becoming close to $0$$0$.

$\int f\left(x\right)dx=c+{\int }_{0}^{x}f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx} = c + {\int}_{0}^{x} f \left(x\right) \mathrm{dx}$

•  The left-hand-side gives the general form of integration. it is later explained as anti-derivative or indefinite integral.

•  The $c$$c$, the constant of integration, is the initial value of the result.

•  The $0$$0$ and $x$$x$ are the starting point and terminal points of the summation.

examples

Finding the integral of $y=7x$$y = 7 x$ in first principles:

$\int 7xdx$$\int 7 x \mathrm{dx}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(\frac{7ix}{n}\right)×\frac{x}{n}$$\left(\frac{7 i x}{n}\right) \times \frac{x}{n}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(\frac{7ix}{n}\right)×\frac{x}{n}$$\left(\frac{7 i x}{n}\right) \times \frac{x}{n}$

$=c+\underset{n\to \infty }{lim}\frac{7{x}^{2}}{{n}^{2}}\sum _{i=1}^{n}i$$= c + {\lim}_{n \to \infty} \frac{7 {x}^{2}}{n} ^ 2 {\sum}_{i = 1}^{n} i$

substituting $\sum _{i=1}^{n}i=n\left(n+1\right)/2$${\sum}_{i = 1}^{n} i = n \left(n + 1\right) / 2$
$=c+\underset{n\to \infty }{lim}\frac{7{x}^{2}}{{n}^{2}}×\frac{n\left(n+1\right)}{2}$$= c + {\lim}_{n \to \infty} \frac{7 {x}^{2}}{n} ^ 2 \times \frac{n \left(n + 1\right)}{2}$

$=c+\underset{n\to \infty }{lim}\frac{7{x}^{2}}{{n}^{2}}×\frac{{n}^{2}+n}{2}$$= c + {\lim}_{n \to \infty} \frac{7 {x}^{2}}{n} ^ 2 \times \frac{{n}^{2} + n}{2}$

$=c+\underset{n\to \infty }{lim}\frac{7{x}^{2}}{2}×\left(1+\frac{1}{n}\right)$$= c + {\lim}_{n \to \infty} \frac{7 {x}^{2}}{2} \times \left(1 + \frac{1}{n}\right)$

applying limit
$=c+\frac{7{x}^{2}}{2}×\left(1+0\right)$$= c + \frac{7 {x}^{2}}{2} \times \left(1 + 0\right)$

$=c+7{x}^{2}/2$$= c + 7 {x}^{2} / 2$

The above proves

The answer is "$\int 7xdx=c+7{x}^{2}/2$$\int 7 x \mathrm{dx} = c + 7 {x}^{2} / 2$"

Finding the integral of $y=\mathrm{sin}x$$y = \sin x$ in first principles:

$\int \mathrm{sin}xdx$$\int \sin x \mathrm{dx}$

$\underset{n\to \infty }{lim}\sum _{i=1}^{n}$${\lim}_{n \to \infty} {\sum}_{i = 1}^{n}$$\left(\mathrm{sin}\left(\frac{ix}{n}\right)\right)×\frac{x}{n}$$\left(\sin \left(\frac{i x}{n}\right)\right) \times \frac{x}{n}$

$=\underset{n\to \infty }{lim}\frac{x}{n}\sum _{i=1}^{n}\mathrm{sin}\left(\frac{ix}{n}\right)$$= {\lim}_{n \to \infty} \frac{x}{n} {\sum}_{i = 1}^{n} \sin \left(\frac{i x}{n}\right)$

Is it possible to simplify this and resolve the indeterminate value $0/0$$0 / 0$ of limit?

Many functions are not easily solved by forward summation as defined by the first principles. We will learn properties of the limit of summation to solve integration of such functions. Then, it can be easily solved based on some property of the limit of summation.

summary

Integral or Integration of a function : The continuous aggregate of the function $f\left(x\right)$$f \left(x\right)$ is defined as.

$\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$

$=c+{\int }_{0}^{x}f\left(x\right)dx$$= c + {\int}_{0}^{x} f \left(x\right) \mathrm{dx}$

$=c+\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(\frac{ix}{n}\right)×\frac{x}{n}$$= c + {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(\frac{i x}{n}\right) \times \frac{x}{n}$

note1: The definite integrals and anti-derivatives or indefinite integrals are introduced in due course.
note2: The summation has many other forms, Riemann, Lebesgue, and Darboux forms, which will be introduced in due course.

Outline