# Help with Antiderivatives

• Sep 25th 2009, 07:53 PM
yoman360
Help with Antiderivatives
find f(x) if $\displaystyle f''(x)=2+cos x, f'(0)=-1, f(\pi/2)=0$

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

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

What did I do incorrectly?
• Sep 25th 2009, 09:26 PM
Prove It
Quote:

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

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

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

What did I do incorrectly?

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

You should have

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

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

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

Now apply the initial conditions and you'll get two equations to solve simultaneously for $\displaystyle C_1$ and $\displaystyle C_2$.
• Sep 26th 2009, 05:15 AM
yoman360
Quote:

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

You should have

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

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

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

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

oops i made a typo it is f'(0)= -1
NOT f(0)=-1
• Sep 26th 2009, 09:57 AM
yoman360
i still need help
• Sep 26th 2009, 10:04 AM
skeeter
Quote:

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

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

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

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.
• Sep 26th 2009, 11:57 AM
yoman360
Quote:

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.