In this page, integration is defined in first principles.

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A car travels at speed `10m//s`. What is the distance traveled in `2` seconds?

- `20m`
- `20m`
- `10m`

The answer is "`20m`". Speed multiplied by time gives the distance traveled in that time.

A car travels at speed `10m//s` in the first `2` seconds, and at speed `15m//s` in the next `3` seconds. What is the total distance traveled?

- distance `=15 xx 3 = 45`m
- distance `= 10 xx 2 + 15 xx 3 = 65` m
- distance `= 10 xx 2 + 15 xx 3 = 65` m

The answer is "distance `= 10 xx 2 + 15 xx 3 = 65` m"

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

distance = speed1 `xx` time1 `+` speed2 `xx` time2 `+...`

Distance is the aggregate of speed. It is generalized as *aggregate of change.*

A car is moving at speed `20`m/sec. What is the distance covered in `t` sec ?

- without a numerical value for `t` the distance traveled cannot be computed.
- without a numerical value for `t` the distance traveled cannot be computed.
- For any value of `t`, the distance traveled is `s=20t` meter

The answer is "For any value of `t`, the distance traveled is `s=20t` meter". *Measurement can be expressed as a function of a variable.*

A car is moving with speed given as a function of time `v=3t^2+2` m/sec. What is the speed at `t=2sec`?

- `14` meter/sec
- `14` meter/sec
- `6` meter/sec

The answer is "`14` meter/sec". Substitute `t=2` in the formula for `v`.

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

`text(distance) = text(speed) xx text(time)`

• Person A does the following: At `t=2`s, the speed is `14`m/sec. So the distance is `14 xx 2 = 28`m.

• Person B does the following: At `t=1`s, the speed is `5`m/sec. And at `t=2`s, the speed is `14`m/sec. So the distance traveled is `5xx1+14xx(2-1) = 19` m.

Which one is correct?

- Person A
- Person B
- Neither of them
- Neither of them

The answer is "Neither of them". The solution is explained in the subsequent questions.

A car is moving with speed given as a function of time `v=3t^2+2` m/sec. We know that

`text(distance) = text(speed) xx text(time)`

One student wants to calculate the distance traveled.

• At `t=0` second, the speed is `=3xx0+2 = 2`m/sec.

• At `t=1` second, the speed is `= 3xx1+2 = 5`m/sec.

• At `t=2` second, the speed is `=3xx4+2 = 14`m/sec.

Speed varies with time. So, consider time intervals `0` to `1`sec and `1`sec to `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<t<1` `=(2+5)/2 = 3.5` m/sec

The speed during time `1<t<2` `=(5+14)/2 = 9.5` m/sec

So distance traveled is `=3.5xx1+9.5xx(2-1) = 13m`

What is calculated in this?

- Approximate value of distance traveled
- Approximate value of distance traveled
- accurate value of distance traveled

The answer is "Approximate value of distance traveled". It is approximate, because the *speed continuously changes with time*, but the calculation approximates to average over `1` second intervals.

A car is moving with speed given as a function of time `v=3t^2+2`. Which of the following is expected?

- displacement is a numerical value
- displacement is a function of time
- displacement is a function of time

The answer is "displacement is a function of time"

A car is moving with speed given as a function of time `v=3t^2+2`. Which of the following is correct?

- Displacement is aggregate of change of speed. So, one has to figure out how to find the aggregate of change of an algebraic expression.
- Displacement is aggregate of change of speed. So, one has to figure out how to find the aggregate of change of an algebraic expression.
- Displacement is defined only when speed during a time period is given. If the speed is given as a function of a variable, displacement cannot be calculated.

The answer is "Displacement is aggregate of change of speed. So, one has to figure out how to find the aggregate of change of an algebraic expression".

A car is moving with speed given as a function of time `v=3t^2+2`. Displacement is aggregate-of-change. Few students are set to find the aggregate of change*students may work these out to understand*

• Person A found aggregate of change, displacement, for `1` second interval. `sum_(i=1,2,3,...)^(t) (3i^2+2)xx1`

• Person B found aggregate of change, displacement, for `2` second interval. `sum_(i=2,4,6,...)^(t) (3i^2+2)xx2`

• Person C found aggregate of change, displacement, for `0.5` second interval. `sum_(i=0.5,1,1.5,...)^(t) (3i^2+2)xx0.5`

What is actually calculated in each?

- approximation of the distance traveled
- approximation of the distance traveled
- continuous-aggregate displacement

The answer is "approximation of the distance traveled"

A car is moving with speed given as a function of time `v=3t^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. What is the distance shown in the odometer?

- approximate distance traveled by the car
- continuous aggregate distance traveled for any given time
- continuous aggregate distance traveled for any given time

The answer is "continuous-aggregate distance traveled for any given time".

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. An odometer can be used to measure the continuous aggregate distance traveled by the car.

The approximate distance can be computed with time interval `delta` is

`sum_(i=delta, 2delta,3delta,...)^(t) (3i^2+2)xx delta`

where `delta=t/n`,

`n` is the number of steps between `0` and `t`

When is the error in the approximation reduced?

- when the number of steps `n` is small
- when the number of steps `n` is large
- when the number of steps `n` is large

The answer is "when the number of steps `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.

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. An odometer can be used to measure the continuous-aggregate distance traveled by the car.

The approximate distance can be computed with time interval `delta` is

`sum_(i=delta, 2delta,3delta,...)^(t) (3i^2+2)xx delta`

where `delta=t/n`,

`n` is the number of steps between `0` and `t`

When is the algebraic expression exactly the continuous-aggregate distance?

- when the number of steps `n` is `oo`
- when the step width `delta` is `0`
- both the above
- both the above

The answer is "both the above". This is a big jump in understanding. The speed is continuously changing and the summation is done continuously (not in steps) when `n` is `oo`.

Generalizing that, for a function `f(x)` the approximate aggregate of change is `sum_(i=delta,2delta,3delta,...)^(x) f(i)xx delta`

where `delta = x/n`

The same can be given as `sum_(i=1,2,3,...)^(n) f((i x)/n)xx x/n`

where `i` takes positive integer values.

The same is simplified as `sum_(i=1)^(n) f((i x)/n)xx x/n`

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. The approximate distance = `sum_(i=1)^(n) (3((i t)/n)^2+2)xx t/n`

When `n=oo`, the expression gives continuous-aggregate distance

`delta=t/n = 0`

`text(distance) ``= sum_(i=1)^(oo) (3xx 0^2+2)xx 0`

`=0+0+0+... oo text( times)`

`=0 xx oo`

What is the value of `0 xx oo`?

- `0`
- `oo`
- indeterminate value `0//0`
- indeterminate value `0//0`

The answer is "indeterminate value `0//0`".

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. The approximate distance = `sum_(i=1)^(n) (3((i t)/n)^2+2)xx t/n`

When `n=oo`, the expression gives continuous-aggregate distance as `0//0`: indeterminate value.

How does one solve a function evaluating to indeterminate value?

- Use Limit of the function as `n` approaching `oo`.
- Use Limit of the function as `n` approaching `oo`.
- there is no method to solve a function evaluating to `0//0`

The answer is "Use Limit of the function as `n` approaching `oo`."

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. The approximate distance `= sum_(i=1)^(n) (3((it)/n)^2+2)xx t/n`

`n` is the number of steps between `0` and `t`

When `n=oo`, the expression gives continuous-aggregate distance

`= sum_(i=1)^(n) (3((it)/n)^2+2)xx t/n |_(n=oo)`

`=sum_(i=1, 2...)^(n) 0`

`=0xxoo = 0/0`

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

`lim_(n->oo) sum_(i=1)^(n) ``(3((it)/n)^2+2)xx t/n`

`= lim_(n->oo) sum_(i=1)^(n) ``(3((it)/n)^2)xx t/n ``+ lim_(n->oo) sum_(i=1)^(n) (2)xx t/n`

`= lim_(n->oo) (3t^3)/n^3 sum_(i=1)^(n) i^2 ``+ lim_(n->oo) (2t)/n sum_(i=1)^(n) 1`* substituting `sum_(i=1)^(n) i^2 = n(2n+1)(n+1)//6`** and `sum_(i=1)^(n) 1 = n`*

`=lim_(n->oo) (3t^3)/n^3 xx (n(2n+1)(n+1))/6 ``+ lim_(n->oo) (2t)/n xx n`

`=lim_(n->oo) (3t^3)/n^3 xx (2n^3 + 3n^2 + n)/6 ``+ lim_(n->oo)(2t)`

`=lim_(n->oo) (3t^3)/6 xx (2 + 3/n + 1/n^2) ``+ lim_(n->oo) 2t`*applying limit *

`=(3t^3)/6 xx (2 + 0 + 0) + 2t`

`=t^3 + 2t`

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

A car is moving with speed given as a function of time `v(t) = 3t^2+2`. The approximate distance `= sum_(i=1)^(n) (3((it)/n)^2+2)xx t/n`

`n` is the number of steps between `0` and `t`

The continuous-aggregate distance traveled `= lim_(n->oo) sum_(i=1)^(n) ``(3((it)/n)^2+2)xx t/n`

Note that the distance traveled is computed starting from time `t=0`, at which point, the initial distance of the car can be non-zero. So, * the distance = initial distance at time `0` + distance traveled between time `0` and `t`*.

For the given problem, the distance `=c ``+ lim_(n->oo) sum_(i=1)^(n) ``(3((it)/n)^2+2)xx t/n`

where `c` is a constant.

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 "aggregate of change 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 aggregate of change relation.)*

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

• The effect is computed as, the continuous aggregate of change: 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.

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 of change or *aggregate-of-change*

Which of the following is a meaning for the word 'integrate'?

- combine one with another to form a whole
- combine one with another to form a whole
- convert a fraction to an integer

The answer is 'combine one with another to form a whole'. The aggregate of change is aptly named integral or integration.

What is the term used to refer "continuous-aggregate" of a function ?

- Pronunciation : Say the answer once

Spelling: Write the answer once

The answer is 'integration'.

Integration in the context of cause-effect pair in continuous aggregate relation:

If the cause is given by `f(x)` then the effect is computed as integration or integral of `f(x)` denoted as

`int f(x) dx`

`= c + lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx x/n`

The second term is understood to be continuous aggregate between `0` and `x`. This is denoted as

`int f(x) dx`

`= c + int_0^x f(x)dx `

`= c + lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx x/n`

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

A quantity `u=f(x)` is related to another quantity `v` such that `v` is the aggregate of change of `u` with respect to `x`. Then,

`v `

`=int u dx `

`=c + int_0^x u dx `

`= c + lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx x/n`

Note that `int u dx` is another quantity `v`, related to the given quantity `u`.

• `f(x)` is called integrand.

• `x` is the variable of integration.

• `c` is the constant of integration.

• in `int_0^x`, the value `0` denotes the start position of integration called "lower limit" of integration.

• in `int_0^x`, the variable `x` denotes the end position of integration called "upper limit" of integration.

Integration is the continuous-aggregate-of-change given by `c + lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx x/n`

• `lim_(n->oo)` represents "continuous" (as against the discrete number of steps between the start and end positions of summation).

• `sum_(i=1)^(n)` represents "summation"

• `x/n` represents the `n` partitions between the start and end positions.

• `f((ix)/n)` represents the value of function in the `i^(text(th))` partition.

• `f((ix)/n)xx x/n` represents the change contributed by the `i^(text(th))` partition

What is aggregate of change of a function called?

- Pronunciation : Say the answer once

Spelling: Write the answer once

The answer is "integration or integral of the function".

Students can connect the notation `int f(x) dx` as

• the small difference in `x` is given as `dx`

• the multiplication by `f(x)` to `dx` implies, the value of function for the small difference is multiplied.

• `int` denotes the sum with limit of `dx` becoming close to `0`.

`int f(x) dx = c + int_0^x f(x)dx`

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

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

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

The aggregate of change of a function with respect to the variable is the integration of the function.

**Integral or Integration of a function** : The aggregate of change of the function `f(x)` is defined as.

`int f(x) dx`

`=c+ int_0^x f(x) dx`

`=c + lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx 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.*

*Solved Exercise Problem: *

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

`int 7x dx`

`= c + lim_(n->oo) sum_(i=1)^(n) ``((7ix)/n)xx x/n`

`= c + lim_(n->oo) sum_(i=1)^(n) ``((7ix)/n)xx x/n `

`= c + lim_(n->oo) (7x^2)/n^2 sum_(i=1)^(n) i `* substituting `sum_(i=1)^(n) i = n(n+1)//2`*

`= c + lim_(n->oo) (7x^2)/n^2 xx (n(n+1))/2`

`= c + lim_(n->oo) (7x^2)/n^2 xx (n^2 + n)/2 `

`= c + lim_(n->oo) (7x^2)/2 xx (1 + 1/n ) `*applying limit *

`= c + (7x^2)/2 xx (1 + 0 )`

`= c + 7x^2//2`

What does the above prove?

- `int 7x dx = c + 7x^2 //2`
- `int 7x dx = c + 7x^2 //2`
- `int 7x dx = c + 0+0`

The answer is "`int 7x dx = c + 7x^2 //2`"

*Solved Exercise Problem: *

Finding the integral of `y=sinx` in first principles:

`int sinx dx`

`lim_(n->oo) sum_(i=1)^(n) ``(sin((ix)/n))xx x/n`

`= lim_(n->oo) x/n sum_(i=1)^(n) sin((ix)/n) `

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

- the summation is not easily simplified
- the summation is not easily simplified
- it is not easily solved in the forward summation
- it can be easily solved based on some property of the limit of summation
- all the above

The answer is "all the above". 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.

*slide-show version coming soon*