Definition of Integral
A derivative of a function
expressed with 3x^{2} + 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




By looking at Table 1 it
appears that if f ‘ (x) = x^{n}, then
However, if we look at
Table 2 it can be seen that the anti differential of 6x comes from various
functions 3x^{2} + 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 antidifferential 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) = 6x^{2} – 2x + 6, and
the function value f (2) = 7. Determine the function formula.
Answers:
f(x) = ∫ f ‘ (x) dx
f(x) = ∫ (6x^{2} –
2x + 6) dx
f(x) = 2x^{3} – x^{2}
+ 6x + c
remember that f(2) = 7 then
f(2) = 2 . 2^{3} 
2^{3} + 6 . 2 + c
7 = 16 – 4 + 12 + c
ó c =  31
So , f(x) = 2x^{3}
– x^{2} + 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 = x^{2} – 4x + c
the point (2, 0) is passed through the curve, then the number pair satisfies the equation
y = x^{2} – 4x + c
0 = 2^{2} – 4 . 2
+ c
ó c = 4
So, the curve equation is y = x^{2} – 4x + 4.
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Nice info
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