# Thread: Variability of functions and their subsequent affects on the formulae for integrals

1. ## Variability of functions and their subsequent affects on the formulae for integrals

Hi all

A table of integrals was provided for us in our text-book:

I have two questions in this regard:

1) What would happen to the antiderivatives, F(x), if a constant factor, b, was inserted infront of all all of the functions, f(x)? For instance instead of f(x) = x^n , it would read b* x^n, instead of f(x)= 1/x, it would read f(x)= b* 1/x and so on and so forth.

2) If f(x)= 1/x = x^-1 corresponds to F(x)= lnx + k (where k is a constan term), then what would happen to the formula for F(x), if the numerator (or exponent, depending on which formulation you go by) in f(x) was replaced by a greater or lesser number? I.e. Instead of f(x)= 1/x, it would read 3/x , 0,5/x or -4/x? And same thing if we go by the second formulation that f(x)= x^-1- what if instead of -1, the exponent was -2, 0,3 or 1/5?

Thank you for any help in advance.

2. These are exactly the same question. Please feel free to move multiplicative constants in and out of the integral at will.

Note: This does not change the arbitrary constant (called 'k' in your chart. If you multiply an arbitrary constant by 3, it's still an arbitrary constant. You can track changes like this of you like, but it's generally a waste of time and energy.

3. Originally Posted by TKHunny
These are exactly the same question.

Hi TKHunny

Hmm... I don't see how my two questions are one and the same? How can adding a factor to a function be the same as changing the value of an exponent in a function?

Please feel free to move multiplicative constants in and out of the integral at will.
I'm not quite sure I understand what you mean. Maybe if you explained by way of examples it would be better.

For instance what would F(x) be if f(x)= K * x^a? Also what would F(x) be, if f(x)= 2/x and if
f(x)= -2/x?

Again, thanks.

4. Originally Posted by Lia85
Hi all

A table of integrals was provided for us in our text-book:

I have two questions in this regard:

1) What would happen to the antiderivatives, F(x), if a constant factor, b, was inserted infront of all all of the functions, f(x)? For instance instead of f(x) = x^n , it would read b* x^n, instead of f(x)= 1/x, it would read f(x)= b* 1/x and so on and so forth.
One of the first things you should have learned about integrals is the
\int f(x)+ g(x) dx= \int f(x)dx+ \int g(x)dx and \int c f(x)dx= c \int f(x)dx for any constant c. Together those say that the integration operator is "linear".

[qpote\ 2) If f(x)= 1/x = x^-1 corresponds to F(x)= lnx + k (where k is a constan term), then what would happen to the formula for F(x), if the numerator (or exponent, depending on which formulation you go by) in f(x) was replaced by a greater or lesser number? I.e. Instead of f(x)= 1/x, it would read 3/x , 0,5/x or -4/x? And same thing if we go by the second formulation that f(x)= x^-1- what if instead of -1, the exponent was -2, 0,3 or 1/5?[/quote]
If you mean multiplied by a number, same thing: \int cf(x)dx= c\int f(x)dx.

The integral \int x^{n}dx, with n\ne -1, is \int x^n dx= \frac{1}{n+1} x^{n+ 1}+ C just as your table says, first row, left. That is because \frac{dx^n}{dx}= nx^{n-1} and integration is the "inverse" of differentiation- to differentiate $x^n$ you multiply by the exponent and subtract 1 from the exponent. Notice that if n= 0, the derivative of x^0 is 0x^{0-1}= 0x^{-1}= 0 which means that there is no anti-derivative for $x^{n-1}$ times a non-zero number. That's why there is a different formula for \int \frac{1}{x}dx= \int x^{-1}dx.

Thank you for any help in advance.