1. ## Antiderivative

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...

2. $f'(x) = \frac{x^{2}}{2} - 2x + C$

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

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

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

now just solve the above two equations simultaneously to find the values of C and D

3. 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)

4. Originally Posted by rushin25
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)
$f'(x) = -\frac{1}{x} + C$

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

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

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

so $C = \frac{\ln 5}{4}$

and $D = - \frac{\ln 5}{4}$

5. Originally Posted by rushin25
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.