# Integration Calculus

Integration is a mathematical concept of adding segements to find the final result, it is also known has a reserve of differetiation. the result from an Integration is an integral

## How to Integrate

Equation: y = ${\mathrm{ax}}^{3}$

1. Add one (1) to the power. In this case: ${\mathrm{ax}}^{3 + 1}$

2. Divide the cooefficient by the new power. In this case: $\frac{\mathrm{{\mathrm{ax}}^{4}}}{4}$

3. add C to the equation which is the constant . In this case $\frac{\mathrm{{\mathrm{ax}}^{4}}}{4}+C$

4. Merge the steps together. $\frac{\mathrm{{\mathrm{ax}}^{4}}}{4}+C$

### Point to note

1. when integrating a number the result contains a variable which is added to the number

The Integration of the differetiated equation above is found by the following steps

Given this equation $y={\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-36x-3$ prove by reversing the differetiated equation $\frac{\mathrm{dy}}{dx}$ back this equation

The differetiated equation $\frac{\mathrm{dy}}{dx}={\mathrm{6x}}^{2}-\mathrm{6x}-36$

Solution:

$\frac{\mathrm{dy}}{dx}={\mathrm{6x}}^{2}-\mathrm{6x}-36$

$\int \left({\mathrm{6x}}^{2}-\mathrm{6x}-36\right)dx$

$\int \left(\frac{{\mathrm{6x}}^{2 + 1}}{3}-\frac{{\mathrm{6x}}^{1 + 1}}{2}-36x+C\right)dx$

$\int \left(\frac{{\mathrm{6x}}^{3}}{3}-\frac{{\mathrm{6x}}^{2}}{2}-36x+C\right)dx$

$\int \left({\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-36x+C\right)dx$

Integral : $\int \left({\mathrm{2x}}^{3}-{\mathrm{3x}}^{2}-36x+C\right)dx$

C is -3 which is the constant of the equation

### Example 1

Integrate the following equations with respect to x

(a) $y={\mathrm{20x}}^{4}$

(b) $y=\mathrm{2x}$

(c) $y={\mathrm{12x}}^{3}+19$

Solutions

(a)

$y={\mathrm{20x}}^{4}$

$\int ={\mathrm{20x}}^{4}$

$\int ={\frac{\mathrm{20x}}{5}}^{4 + 1}$

$\int ={\frac{\mathrm{20x}}{5}}^{5}$

Answer: $\int ={\mathrm{4x}}^{5}$ + C

or $\int ={\mathrm{4x}}^{5}$

(b)

$y=\mathrm{2x}$

$\int =\mathrm{2x}$

$\int ={\frac{\mathrm{2x}}{2}}^{1 + 1}$

$\int ={\frac{\mathrm{2x}}{2}}^{2}$

Answer: $\int ={x}^{2}$ + C

or $\int ={x}^{2}$

(c)

$y={\mathrm{12x}}^{3}+19$

$\int ={\mathrm{12x}}^{3}+19$

$\int ={\frac{\mathrm{12x}}{4}}^{3 + 1}+19x$

$\int ={\frac{\mathrm{12x}}{4}}^{4}+19x$

Answer: $\int ={\mathrm{3x}}^{4}+19x$ + C

or $\int ={\mathrm{3x}}^{4}+19x$

### Example 2

Evaluate $∫ -1 2 ( 2 + x - x 2 ) dx$

$[ 2x + x 2 2 - x 3 3 ]$

$[ 2(2) + 2 2 2 - 2 3 3 ]$ - $[ 2(-1) + -1 2 2 - -1 3 3 ]$

$[ 4 + 2 - 8 3 ]$ - $[ -2 + 1 2 - -1 3 ]$

$[ 6 - 2.67 ]$ - $[ -2 + 0.5 + 0.33 ]$

$[ 3.33 ]$ - $[ -1.17 ]$

$[ 3.33 ]$ + $[ 1.17 ]$

Answers: $4.5$

## How to obtaining an integral from a derivation

From this quadratic equation ${x}^{2}+2x-3=0$ obtain a derivative and then reserve it to a quadratic equation by obtaining an Integral

$\frac{\mathrm{Dy}}{Dx}={x}^{2}+\mathrm{2x}-3$

$={\mathrm{2x}}^{1}+{\mathrm{2x}}^{1-1}-0$

$={\mathrm{2x}}^{1}+{\mathrm{2x}}^{0}-0$

Derivative: $\frac{\mathrm{Dy}}{Dx}=\mathrm{2x}+2$

$\int =\left(\mathrm{2x}+2\right)dx$

$\int =\left(\frac{{\mathrm{2x}}^{1+1}}{2}+\mathrm{2x}+C\right)dx$

$\int =\left(\frac{{\mathrm{2x}}^{2}}{2}+\mathrm{2x}+C\right)dx$

$\int =\left({x}^{2}+\mathrm{2x}+C\right)dx$

Integral: $\int =\left({x}^{2}+\mathrm{2x}+C\right)$

## Integral with limits

Evaluate $∫ 1 3 ( 3x 2 + 4x ) dx$

$[ 3(3) 2 + 4(3) ] - [ 3(1) 2 + 4(1) ]$

$[ 3(9) + 12 ] - [ 3(1) + 4(1) ]$

$[ 27 + 12 ] - [ 3 + 4 ]$

$[ 39 ] - [ 7 ]$

Answer: $32$