In this page, indefinite integral and definite integrals are explained. In definite integrals, various forms of limit of summation are introduced.

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Summary: Integral of a function `f(x)` is specified with an initial value or the constant of integration `c`.

Integral of `f(x)` with respect to variable `x` `int f(x) dx =int_0^x f(x)dx + c`

The LHS is referred as indefinite integral.

The integration is taken till a variable `x`, so the integral is called indefinite.

Contrasting this with "definite integral" : If the integration is taken in a defined interval `[a,b]`, then, the integral `int_a^b f(x) dx` is definite integral.

Either of the integration can be loosely referred as "area function".

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

- clearly stated with exact boundaries
- clearly stated with exact boundaries
- countable and less in numbers

The answer is 'clearly stated with exact boundaries'.

What is the term used to refer the integral taken for a closed interval `[a,b]` : given as `int_a^b f(x) dx` ?

- Pronunciation : Say the answer once

Spelling: Write the answer once

The answer is 'definite integral'.

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

- of unknown boundary that can change
- of unknown boundary that can change
- uncountable and many in numbers

The answer is 'of unknown boundary that can change'.

What is the term used to refer the integral for a variable interval `[0,x]` with an initial constant `c`, given as `int f(x) dx = int_0^x f(x)dx + c`?

- Pronunciation : Say the answer once

Spelling: Write the answer once

The answer is 'indefinite integral'.

definite integral is the aggregate of change between defined start and end values.

indefinite integral is the aggregate of change to a variable end value with a constant as initial value.

**Definite Integral: **For a given function, continuous and integrable in interval `[a,b]`, the definite integral is `int_a^b f(x) dx` `= lim_(n->oo) sum_(i=1)^n f(i(b-a)/n) xx (b-a)/n` **Indefinite Integral: **For a given function continuous and integrable, the definite integral is `int f(x) dx` `=c+ lim_(n->oo) sum_(i=1)^n f((ix)/n) xx x/n`

Note: Indefinite integral is also called anti-derivatives. This will be explained in due course.

Summary of integration: Given `y=f(x)` a function of variable `x`, the integral of `y` is

`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` We understood that the summation is given by vertical bars under the curve, as depicted in the figure.

Which of the following is correct?

- the definition of "limit of summation" is unique
- alternate forms of "limit of summation" are possible, as long as the continuous aggregate is represented
- alternate forms of "limit of summation" are possible, as long as the continuous aggregate is represented

The answer is "alternate forms of "limit of summation" are possible, as long as the continuous aggregate is represented".

The continuous aggregate is represented under conditions

• the entire interval between `0` and `x` is covered

• the width of all partitions (of interval between `0` and `x`) tends to `0` as `n` tends to infinity.

An example for an alternate representation is

`int f(x)dx`

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

where `r<1`

This is represented in the figure. As `n` tends to infinity, all the partition widths are `0` and the partition area is identical to the area under the curve.

Consider `y=x^k`. The integral represented as

`int f(x)dx`

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

There is no simple method to find the limit of this mathematical formulation. So an alternate method is used.

`int f(x)dx`

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

where `r<1`

In both the representations, as `n` tends to infinity, all the partition widths are `0` and the partition area is identical to the area under the curve.*Note: When it was not possible to work-out the limit of the summation, it is about the mathematical formulation of the integral, not the integrability of the function. If the limit of summation is defined, then the function is integrable.*

To understand using different forms of summation, let us learn the Fermat's method to calculate the integral of `f(x) = x^k`.

For simplicity, the constant `c` is taken to be `0`, and the integral is between `0` and `x`. `int_0^x f(x)dx`

`=lim_(n->oo) sum_(i=0)^n ``f(xr^i) xx (xr^i - xr^(i+1))`

`=lim_(n->oo) sum_(i=0)^n ``x^k r^(ki) xx xr^i(1-r)`

`=lim_(n->oo) sum_(i=0)^n ``x^(k+1) r^((k+1)i) (1-r)`

`=lim_(n->oo) x^(k+1) ``(1-r) sum_(i=0)^n r^((k+1)i)` * using geometrical series for sum `sum_(i=0)^(oo) r^((k+1)i) ``= 1//[(1-r)(1+r+r^2...r^n)]`*

`=lim_(n->oo) ``x^(k+1) (1-r) ``//[(1-r)(1+r+r^2...r^k)] `

`=lim_(n->oo) ``x^(k+1)/(1+r+r^2...r^k) ` * as limit n tends to `oo`, the `r` tends to `1`.*

`=x^(k+1)/(1+1+1...(k+1) text(times)) `

`=x^(k+1)/(k+1)`

What does the above prove?

- `int x^k dx = x^(k+1)/(k+1) + c`
- `int x^k dx = x^(k+1)/(k+1) + c`
- `int x^k dx = (k+1) + c`

The answer is "`int x^k dx = x^(k+1)/(k+1) + c`"

We understood that the continuous aggregate is represented under conditions

• the whole interval between `0` and `x` is covered

• the width of all partitions (of interval between `0` and `x`) tends to `0` as `n` tends to infinity.

Definition of Riemann integral is illustrated and explained.*Note: This explains without going into rigorous mathematical formulations.* For a given function `f(x)` continuous and real valued in a closed interval `[a,b]`, the definite integral `int_a^b f(x)dx` is illustrated in the figure.

• the partitions can be of different widths

• the partitions span the interval `[a,b]`

• the height of the partitions is any one value of `f(x)` in that partition width.

• limit on the number of partitions tending to `oo` and thus the partition widths tends to `0`.

Under the limit, if the sum is defined, then the function is Riemann integrable in that interval.

When `n` tends to infinity, will this integral be identical to the one defined with partitions of equal widths?

- Yes
- Yes
- No

The answer is "Yes"

Definition of Darboux integral is illustrated and explained.*Note: This explains without going into rigorous mathematical formulations* For a given function `f(x)` continuous and real valued in a closed interval `[a,b]`, the definite integral `int_a^b f(x)dx` is illustrated in the figure.

• the partitions can be of unequal widths

• the partitions span the interval `[a,b]`

Two estimates are calculated (1) lower estimate and (2) higher estimate.

• the height of the partitions of lower estimates are the minimum values of `f(x)` in that partition.

• the height of the partitions of higher estimates are the maximum value of `f(x)` in that partition.

• limit on the number of partitions tending to `oo` and thus the partition width tending to `0`.

• under the limit, if the lower estimate and higher estimate are equal, then the function is Darboux integrable in that interval.

Will the result of Darboux integral and Riemann integral be same?

- Yes, for integrable functions the two are identical.
- Yes, for integrable functions the two are identical.
- No. The two results differ.

The answer is "Yes, for most integrable functions the two are identical."

Definition of Lebesgue integral is illustrated and explained.*Note: This explains without going into rigorous mathematical formulations* For a given function `f(x)` continuous and real valued in a closed interval `[a,b]`, the definite integral `int_a^b f(x)dx` is illustrated in the figure.

• the partitions are horizontal as illustrated.

• the partitions can be of unequal heights

• the horizontal partitions span the area under the curve

• limit on the number of partitions tending to `oo` and thus the partition height tends to `0`.

• Under the limit, if the sum is defined, then the function is Lebesgue integrable in that interval.

Will the result of Lebesgue integral and Riemann integral be same?

- Yes, for integrable functions the two are identical.
- Yes, for integrable functions the two are identical.
- No. The two results differ.

The answer is "Yes, for integrable functions the two are identical"

integral, as aggregate of change, can be defined in different forms.

**Different forms of integral: **

• Riemann Integral

• Darboux Integral

• Lebesgue Integral

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