
#1
Dec503, 03:33 AM

P: 40

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. 



#2
Dec503, 05:06 AM

P: 261

1.The simplest way to solve it is to remember what is the derivative of 2^{x},by integrating the known equality.
(In the general case [a^{x}]'=a^{x}*lna with a=const) 2.Let 2^{x}=t x=(1/ln2)*lnt > dx=(1/ln2)*1/t*dt Further is straightforward. 



#3
Dec503, 07:16 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

One way to do this is to note that, since e^{x} and ln(x) are inverse functions, x= e^{ln(x)} for all x.
In particular, 2^{x}= e^(ln(2^{x})= e^{x ln(2)} so that d(2^{x})/dx= de^{x ln(2)}/dx= ln(2) 2^{x}. (I'll bet that derivative formula is somewhere in your text.) Since d(2^{x})/dx= ln(2) 2^{x}, the antiderivative of 2^{x} is (1/ln(2)) 2^{x}. In general, the derivative of a^{x} is ln(a) a^{x} and the antiderivative is (1/ln(a)) a^{x}. (Notice that if a= e, ln(e)= 1 and we get the standard formulas.) 


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