# antiderivatives

#### rushin25

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
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 $$\displaystyle C$$.

#### rushin25

but what will be the value for 4(pi/2)^2 ?

#### Prove It

MHF Helper
but what will be the value for 4(pi/2)^2 ?
Surely $$\displaystyle 4\left(\frac{\pi}{2}\right)^2 = 4\left(\frac{\pi^2}{4}\right) = \pi^2$$...

#### rushin25

lol...my bad...didn't see that... so should i use the pi value in radian?

#### Prove It

MHF Helper
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.