antiderivatives

Oct 2009
8
0
f ''(x) = 8 + cos(x)

f(0) = -1


here is what i got
f '(x)= 8x + sinx +C
f(x) = 4x^2 - cos x + Cx +D

since f(0)= -1
f(0)= 0-1+0+D= -1
D = 0

Now the problem is..how do i substitute the value
in f(x) ???
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
f ''(x) = 8 + cos(x)

f(0) = -1


here is what i got
f '(x)= 8x + sinx +C
f(x) = 4x^2 - cos x + Cx +D

since f(0)= -1
f(0)= 0-1+0+D= -1
D = 0

Now the problem is..how do i substitute the value
in f(x) ???
If \(\displaystyle D = 0\)

then \(\displaystyle f(x) = 4x^2 - \cos{x} + Cx\).


So now

\(\displaystyle 0 = 4\left(\frac{\pi}{2}\right)^2 - \cos{\left(\frac{\pi}{2}\right)} + \frac{\pi}{2}C\).

Simplify and solve for \(\displaystyle C\).
 
Oct 2009
8
0
lol...my bad...didn't see that... so should i use the pi value in radian?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
lol...my bad...didn't see that... so should i use the pi value in radian?
Yes, if you are given values of \(\displaystyle \pi\) then it is safe to assume you are measuring in radians.