Antiderivative

Oct 2009
8
0
Find f. (x>0) f ''(x) = x^-2
x > 0
f(1) = 0
f(5) = 0

i am having trouble finding f(x) because the denominator goes to infinity...:(
 
Last edited:
May 2009
959
362
\(\displaystyle f'(x) = \frac{x^{2}}{2} - 2x + C \)

\(\displaystyle f(x) = \frac{x^{3}}{6} - x^{2} + Cx + D \)

\(\displaystyle f(1) = 0 = \frac{1}{6} -1 + C + D \)

\(\displaystyle f(5) = 0 = \frac{125}{6} - 25 + 5C + D \)

now just solve the above two equations simultaneously to find the values of C and D
 
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Oct 2009
8
0
i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)
 
May 2009
959
362
i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)
\(\displaystyle f'(x) = -\frac{1}{x} + C \)

\(\displaystyle f(x) = - \ln x + Cx + D \)


\(\displaystyle f(1) = 0 = -\ln 1 + C + D = C + D\)

\(\displaystyle f(5) = 0 = -\ln 5 + 5C + D \)


so \(\displaystyle C = \frac{\ln 5}{4} \)

and \(\displaystyle D = - \frac{\ln 5}{4} \)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
i m sorry .... i guess i made a mistake while posting the question...it is

f "(x)= x^-2
x > 0
f(1) = 0
f(5) = 0

find f(x)
It goes to infinity at x= 0 but that has nothing to do with this problem. It says explicitely (in your first post) that you are to find f(x) for x> 0.