In this page, the integrals of functions that are given as arithmetic operation of multiple functions is discussed. The arithmetic operations are multiplication by a constant, addition, and subtraction.

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Indefinite integral or anti-derivative is defined as `int f(x)dx = g(x) +c`, where

• `g(x)` is found by aggregate-of-change represented by `lim_(n->oo) sum_(i=1)^n f((i x)/n) xx x/n`

• `g(x)` is found such that `d/(dx) g(x) = f(x)`.

• `g(x)` is the area under the curve `f(x)` between `0` and `x`. To understand the properties of indefinite integrals one of these is used.

• aggregate of change provides rigorous mathematical proof.

• anti-derivatives provide proof derived from properties of derivatives.

• area under the curve provide the geometrical methods. *A proof given in one of these three can be verified in other. *

Integral of a scalar multiple of a function: Given `v(x)=color(deepskyblue)au(x)`.

`int v dx `

`quad quad = lim_(n->oo) sum_(i=1)^n (v((i x)/n) x/n`

`quad quad = lim_(n->oo) sum_(i=1)^n (color(deepskyblue)au((i x)/n) x/n`* with continuity and integrability conditions on `u`*

`quad quad = color(deepskyblue)a lim_(n->oo) sum_(i=1)^n (u((i x)/n) x/n`

`quad quad = color(deepskyblue)a int u dx`

what does the above prove?

- `int a v dx = a int v dx `
- integral of a multiple of a function is multiple of the integral of the function
- Both the above
- Both the above

The answer is "both the above"

Intuitive understanding for

`int au dx; = a int u dx`

• aggregate of change multiplies when the function is multiplied by a constant. • area under the curve multiplies when the `y` values of curve is multiplied.

*Solved Exercise Problem: *

Given `int y dx = 2x^2` and `v=y/5`, what is `int v dx`?

- `(2x^3)/3`
- `2/5 x^2`
- `2/5 x^2`

The answer is "`2/5 x^2`"

`int v dx`

`= int y/5 dx`

`= 1/5 int y dx`

`=1/5 xx 2x^2`

`=2/5 x^2`

Finding integral of sum or difference.

`int (u+v)dx`

`quad = lim_(n->oo) sum_(i=1)^n ``(u((i x)/n)+-v((i x)/n))``xx x/n`

`quad = lim_(n->oo) sum_(i=1)^n ``u((i x)/n)xx x/n +- v((i x)/n)``xx x/n`* with continuity and integrability conditions on `u` and `v`*

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

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

`quad quad = int u dx +- int v dx`

What does the above prove? `

- `int (u+v) dx = int u dx + int v dx`
- `int (u+v) dx = int u dx + int v dx`
- integral of a sum or difference is the sum or difference of integrals.
- `int (u-v) dx = int u dx - int v dx`
- all the above

The answer is "all the above".

Intuitive understanding for

`int (u+-v) dx = int u dx +- int v dx`

• aggregate of change adds (subtracts) when the function is added(subtracted). • Areas under the curves add (or subtract) when the functions are added (or subtracted).

*Solved Exercise Problem: *

Given `int u dx = sinx` and `int v dx =x+20`, what is `int (u+v) dx`?

- `(sin x)(x+20)`
- `sin x + x + 20`
- `sin x + x + 20`

The answer is "`sin x + x + 20`"

Integral of a multiple of a function is multiple of the integral of the function.

Integral of a sum or difference is sum or difference of integrals.

Note: Other algebraic operations (like multiplication, function of function, etc.) will be taken up in due course.

**Integral of a Multiple: **

`int (au)dx = a int u dx`**Integral of Sum or Difference: **

`int (u+v) dx = int u dx + int v dx`

`int (u-v) dx = int u dx - int v dx`

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