steffen ecca said:
Homework Statement
Waht is the correct integral of
Homework Equations
2^{-x}
The Attempt at a Solution
Is -{1/(x-1)}*2^{1-x} correct or do I have to apply some substitution?
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
