Find f(e) if f'(x) = 10^x + 1/2x , f(1) = 2
+C dont forget the constant solve for the constant by using the information f(1) = 2 then determine f(e)
didnt check if the integration is correct
useful tool for checking integrations http://integrals.wolfram.com/index.jsp?expr=10^x%2B1%2F%282x%29&random=false however i wouldnt suggest relying on it too heavily
did I even do the antiderivative correctly..?
sadly, you did not. the antidervative of a^x is a^x/ln a
can't u take the ln of it?
and then product rule..
and for 1/2x , do you separate them as 1/2 x 1/x , then antiderivative of 1/x is ln x
if you take ln, then you have to apply e to keep it equivalent to to original, like this one \[10^x =e^{\ln 10^x}\]
you are right on an antiderivative of the second term 1/(2x)
ohh ok thanks. but if u ln then e, they give the same results as 10^x ln 10?
x ln 10 = ln10 + 1/ 10 x < if u apply e , it will be 10 + e^1/10 x which doesnt realy make sense
\[\ln (10^x)=x \ln 10\] and you don't have to take the derivative. it is already the derivative of the f(x)
ohh i get it thx. when i do the next step 2= 10ln10 + 1/2 ln 2 +C ln 10 + ln2 , do i combine to ln (10x2)
can' isolate ln
uh, i'm having slow refresh right now
i'll juz wait :)
i doubt you can combine them i suggest grabbing a calcuator and calculating for C
my answer need to be in exact numbers
if thats the case, then definitely use a calculator
oh actually nvm, i think i know how, i've put the wrong numbers in
your function should be \[f(x)=\frac{ 10^x }{ \ln 10 }+\frac{ 1 }{ 2 } \ln x+C\]
ummm.i'm also trying to do it with xln10, but when it asks for f(e), it is weird to have e ln 10
why is ln in the denominator, isnt a^x is a^x ln x?
that is for differentiation. the antidiferrentiation causes it to be "under"
so x ln 10 is not the antiderivative too?
it is not. is is 10^x / ln 10
just ignore ln then e process i said earlier. it only confused you.
ohhh ok , yes ok, got it! thanks again sirm :D u helped alot!
yaw (you're always welcome) :D
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