1. ## Help with Antiderivatives

find f(x) if $f''(x)=2+cos x, f'(0)=-1, f(\pi/2)=0$

This is what I get:
$f'(x)=2x+sin x -1$
$f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2$

The correct answer is: $x^2-cos x-\frac{1}{2}\pi x$

What did I do incorrectly?

2. Originally Posted by yoman360
find f(x) $f''(x)=2+cos x, f(0)=-1, f(\pi/2)=0$

This is what I get:
$f'(x)=2x+sin x -1$
$f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2$

The correct answer is: $x^2-cos x-\frac{1}{2}\pi x$

What did I do incorrectly?
I don't know how you got $f'(x) = 2x + \sin{x} - 1$, because your two initial conditions are for $f$, not $f'$.

You should have

$f''(x) = 2 + \cos{x}$

$f'(x) = 2x + \sin{x} + C_1$

$f(x) = x^2 - \cos{x} + C_1x + C_2$.

Now apply the initial conditions and you'll get two equations to solve simultaneously for $C_1$ and $C_2$.

3. Originally Posted by Prove It
I don't know how you got $f'(x) = 2x + \sin{x} - 1$, because your two initial conditions are for $f$, not $f'$.

You should have

$f''(x) = 2 + \cos{x}$

$f'(x) = 2x + \sin{x} + C_1$

$f(x) = x^2 - \cos{x} + C_1x + C_2$.

Now apply the initial conditions and you'll get two equations to solve simultaneously for $C_1$ and $C_2$.
oops i made a typo it is f'(0)= -1
NOT f(0)=-1

4. i still need help

5. Originally Posted by yoman360
find f(x) if $f''(x)=2+cos x, f'(0)=-1, f(\pi/2)=0$

This is what I get:
$f'(x)=2x+sin x -1$
$f(x)=x^2-cos x-x+(\frac{\pi}{2})-(\frac{\pi}{2})^2$

The correct answer is: $x^2-cos x-\frac{1}{2}\pi x$

What did I do incorrectly?
based on the equation for f''(x) and the given initial conditions, the "correct" answer you posted is incorrect.

your solution for f(x) is correct.

6. Originally Posted by skeeter
based on the equation for f''(x) and the given initial conditions, the "correct" answer you posted is incorrect.

your solution for f(x) is correct.
ok thanks i tried this problem numerous times and i keep getting my answer which is the correct answer i guess.