# Definition of Integral

A derivative of a function expressed with 3x2 + 7x + 5 is 6x + 7. If it is reversed, can you determine the formula of a function if it is known that the function of the decline is 6x + 7? The process of determining a function if the derivative is known is what is called Anti Differential or Integral. So, integral is the opposite of differential. To determine the integral of a function, it is not as easy as time to look for derivative functions. In order for you to get a clear picture, consider the derivatives of several functions as shown in the table below,

Table 1
Table 2
 f(x) f ’(x) x 1 1/2 x2 x 1/3 x3 x2 1/4 x4 x3 1/5 x5 x4
 f(x) f ’(x) 3x2 + 200 6x 3x2 + 300 6x 3x2 – 5 6x 3x2 – 300 6x 3x2 + 1/4 6x

By looking at Table 1 it appears that if f ‘ (x) = xn, then

However, if we look at Table 2 it can be seen that the anti differential of 6x comes from various functions 3x2 + c, with c a constant. From the various things contained in Tables 1 and 2, a rule can be obtained as follows,

### Uncertainty of an Integral

If dy/dx is a notation for derivatives, then the notation for integrals is . For example, a function f(x) is neutralized against x and is written as follows.

∫ f(x) dx.

Read: Integral of f(x) to x

If F(x) anti derivative of f(x), then F(x) + c is also anti derivative of f(x), c is a constant. In general the f(x) integral to x can be written:

∫ f(x) dx = F(x) + c

Rules (*) if written in integral notation are like the following,

The formula above is called the indeterminate integral formula.

Example 1.

### Usage of Uncertain Integral

It has been discussed previously that in determining anti-differential in a derivative function that each contains a value of c (constant) that has not been determined. If we want to determine the function f of a derivative function, then there must be other data so that we can determine the value of c, which is the value of the function in question.

Example 2.
It is known that the derivative of function f is expressed as f ‘ (x) = 6x2 – 2x + 6, and the function value f (2) = -7. Determine the function formula.

f(x) = ∫ f ‘ (x) dx
f(x) = ∫ (6x2 – 2x + 6) dx
f(x) = 2x3 – x2 + 6x + c

remember that f(2) = -7 then

f(2) = 2 . 23 - 23 + 6 . 2  + c
-7   = 16 – 4 + 12 + c
ó  c = - 31
So , f(x) = 2x3 – x2 + 6x – 31.

Example 3.
A curve y = f(x) through a point (2, 0). If the gradient equation is dy/dx = 2x - 4. Determine the equation of the curve,

dy/dx = 2x – 4
y = ∫ (2x – 4) dx
y = x2 – 4x + c

the point (2, 0) is passed through the curve, then the number pair satisfies the equation

y = x2 – 4x + c
0 = 22 – 4 . 2 + c
ó c = 4

So, the curve equation is y = x2 – 4x + 4.