|Dec5-03, 03:33 AM||#1|
Could someone please help me with the integral of 2^x. dx
I bet its really simple but i have looked in several books and they just give the answer.
|Dec5-03, 05:06 AM||#2|
1.The simplest way to solve it is to remember what is the derivative of 2x,by integrating the known equality.
(In the general case [ax]'=ax*lna with a=const)
x=(1/ln2)*lnt ---> dx=(1/ln2)*1/t*dt
Further is straightforward.
|Dec5-03, 07:16 AM||#3|
One way to do this is to note that, since ex and ln(x) are inverse functions, x= eln(x) for all x.
In particular, 2x= e^(ln(2x)= ex ln(2)
so that d(2x)/dx= dex ln(2)/dx= ln(2) 2x. (I'll bet that derivative formula is somewhere in your text.)
Since d(2x)/dx= ln(2) 2x,
the anti-derivative of 2x is (1/ln(2)) 2x.
In general, the derivative of ax is ln(a) ax and the anti-derivative is (1/ln(a)) ax.
(Notice that if a= e, ln(e)= 1 and we get the standard formulas.)
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