# Integration: The Primitive Function

Although Trapezoidal rule and Simpson's rule gave us insight to measure an irregular piece of shape's area, there is still lack of definitive accuracy. Mathematicians researched ways to find the ultimate solution to find irregular areas, and somehow found that there is a link between differentiation and integration. Integration is the link that mathematicians found to calculate the areas under a curve - and integration produces the primitive function . This made possible a simple method for finding exact areas. In fact, differentiation and integration are inverse operations, a discovery that is later call the fundamental theorem of calculus.

## Fundamental Theorem of Calculus

Given a curve with equation y = f(x), the area of the curve could be calculated by means of integration of the equation to obtain its primitive function.

The area is given by:

$\int _{ a }^{ b }{ f(x)dx=F(b)-F(a) }$ where F(x) is the primitive of function f(x)

## Definite Integrals

By the fundamental theorem of calculus, the primitive function of ${ x }^{ n }$ is $\cfrac { { x }^{ n+1 } }{ n+1 } +C$, where a and b are the boundaries and C is a constant.

Using this formula of the primitive function, we can find the area under simple curves.

$\int _{ a }^{ b }{ { x }^{ n } } ={ \left[ \cfrac { { x }^{ n+1 } }{ n+1 } \right] }_{ a }^{ b }\\ \\ \\ =\quad \cfrac { { b }^{ n+1 } }{ n+1 } -\cfrac { { a }^{ n+1 } }{ n+1 }$

### Examples

1. Evaluate $\int _{ 3 }^{ 4 }{ (2x+1)dx }$.

The question is asking the area under the line y = 2x + 1, between the boundaries of x = 3 and x = 4. Therefore we use the integration technique to evaluate its area.

$\int _{ 3 }^{ 4 }{ (2x+1)dx } \\ \\ =\quad { \left[ x^{ 2 }+x \right] }_{ 3 }^{ 4 }\\ \\ =\quad ({ 4 }^{ 2 }+4)-({ 3 }^{ 2 }+3)\\ \\ =\quad 20-12\\ \\ =\quad 8$

To understand the area under the line, refer to the diagram below:

2. Evaluate $\int _{ 0 }^{ 2 }{ -3{ x }^{ 2 }dx }$.

$=\quad -{ \left[ { x }^{ 3 } \right] }_{ 0 }^{ 2 }\qquad <-\quad notice\quad that\quad the\quad -1\quad is\quad a\quad constant\quad outside\quad the\quad integration\\ \\ =\quad -\left[ ({ 2 }^{ 3 })-({ 0 }^{ 3 }) \right] \\ \\ =\quad -(20-12)\\ \\ =\quad -8$

As you have probably noticed that the area is -8 sq. units.

Think: why is the area negative?

## Indefinite Integrals

If the function is not bounded by any region (no lower or upper limit), you can find a general or indefinite integral.

$\int { f\left( x \right) dx } =F(x)+C$ where F(x) is a primitive function of f(x) and C is a constant.

Therefore, the general integration result could be obtained by:

$\int { { x }^{ n }dx } =\cfrac { { x }^{ n+1 } }{ n+1 } +C$

### Examples

Find each indefinite primitive function:

1. $\int { ({ x }^{ 4 }-{ 3x }^{ 2 }+{ 4x }-7)\quad \quad dx }$

$= \quad \cfrac { { x }^{ 5 } }{ 5 } -3\left( \cfrac { { x }^{ 3 } }{ 3 } \right) +4\left( \cfrac { { x }^{ 2 } }{ 2 } \right) -7x+C\\ =\quad \cfrac { { x }^{ 5 } }{ 5 } -{ x }^{ 3 }+2{ x }^{ 2 }-7x+C$

2. $\int { \cfrac { 2{ x }^{ 5 }+7{ x }^{ 2 }-3x }{ x } \quad dx }$

$=\int { \cfrac { 2{ x }^{ 5 } }{ x } +\cfrac { 7{ x }^{ 2 } }{ x } -\cfrac { 3x }{ x } \quad dx } \\ \\ =\int { 2{ x }^{ 4 }+7{ x }-3\quad dx } \\ \\ =\int { \cfrac { 2{ x }^{ 5 } }{ 5 } +\cfrac { 7{ x }^{ 2 } }{ 2 } -3x+C }$

3. $\int { \left( \cfrac { 1 }{ { x }^{ 3 } } +\sqrt { x } \right) \quad dx }$

$=\quad \int { { x }^{ -3 }+{ x }^{ \frac { 1 }{ 2 } }\quad dx } \\ \\ =\quad \cfrac { { x }^{ -2 } }{ -2 } +\cfrac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +C\\ \\ =\quad -\cfrac { 1 }{ 2{ x }^{ 2 } } +\frac { 2\sqrt { { x }^{ 3 } } }{ 3 } +C$

## Function of a Function Rule

Another rule in integration is as follows:

You can use this rule if the power (exponent) of the function is over the whole function (with brackets).

### Examples

1. $\int { { (5x-9) }^{ 3 }\quad dx }$

$=\quad \cfrac { { (5x-9) }^{ 4 } }{ 5\times 4 } +C\\ \\ =\quad \cfrac { { (5x-9) }^{ 4 } }{ 20 } +C$

2. $\int { \sqrt { 4x+3 } \quad dx }$

$=\quad \int { { \left( 4x+3 \right) }^{ \frac { 1 }{ 2 } }\quad dx } \\ \\ =\quad \cfrac { { (4x+3 })^{ \frac { 3 }{ 2 } } }{ 4\times \frac { 3 }{ 2 } } +C\\ \\ =\cfrac { \sqrt { { (4x+3) }^{ 3 } } }{ 6 } +C$