# Integration of 2^{-x}

1. Dec 12, 2012

### steffen ecca

1. The problem statement, all variables and given/known data

Waht is the correct integral of

2. Relevant equations

2^{-x}

3. The attempt at a solution

Is -{1/(x-1)}*2^{1-x} correct or do I have to apply some substitution?

2. Dec 12, 2012

### Simon Bridge

Welcome to PF;
Hint: You can check your solution by differentiating it.
If it comes back with what you started with, you got it right.

It also helps if you state your reasoning - it looks like you've tried the rule for integrating powers of the variable.
Here, the variable is the power, does that make sense?

3. Dec 12, 2012

### Staff: Mentor

Try this:

2 = e ^(ln 2)

ln2 is defined as "the power you have to raise e to to get the number 2"

So the number 2 is equal to e raised to "the power you have to raise e to to get the number 2"

4. Dec 12, 2012

### HallsofIvy

Staff Emeritus
You need to learn very quickly and very thoroughly that the derivative formula $d(x^n)/dx= nx^{n-1}$ and the corresponding integral formula $\int x^n dx= 1/(n+1) x^{n+1}+ C$ only hold when the variable, x, is the base and the exponent is a constant. The situation in which the base is a constant and the exponent is x, is completely different. Both derivative and integral, for example, of $e^x$ is just $e^x$ itself (plus "C" for the integral, of course).

To differentiate something like $e^{f(x)}$, you need to use the chain rule:
Let u= f(x) so that $e^{f(x)}= e^u$. Then $d(e^u)/dx= e^u du/dx$ or $d(e^{f(x)}/dx= e^{f(x)}df/dx$. The integral is harder- we cannot use "substitution", which is essentially the "inverse" of the chain rule, unless we already have the "df/dx= f'(x)" in the integral: $\int e^{f(x)}f'(x)dx= e^{f(x)}+ C$.

In the special case that f(x) is "linear", that is $f(x)= ax+ b$ (like your example here, a= -1, b= 0) then f'(x)= a, a constant, and we can take a constant in and out of an integral at will. To integrate $\int e^{-x}dx$, let u= -x so that du= -dx and now there are two ways of thinking:
1) Multiply by (-1)(-1)= 1, taking one -1 inside the integral: $\int e^{-x}dx= -\int -e^{-x}dx$ and now let u= -x so that du= -dx and $-\int -e^{-x}dx= -\int e^{-x}(-dx)= -\int e^u du$$- e^u+ C= -e^{-x}+ C$.

2) Start of by letting u= -x so that du= -dx and -du= dx so that $\int e^{-x}dx= \int e^u(-du)= -\int e^u du= e^u+ C= e^{-x}+ C$

5. Dec 12, 2012

### Ray Vickson

You should never ask, in this type of question, whether a proposed solution is correct; you should always check by taking the derivative to see whether you get back to the original function. Try it!