Help with Antiderivatives

**yoman360** · September 25th 2009, 07:53 PM

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?

**Prove It** · September 25th 2009, 09:26 PM

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 $C_2$ .

**yoman360** · September 26th 2009, 05:15 AM

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

**yoman360** · September 26th 2009, 09:57 AM

i still need help

**skeeter** · September 26th 2009, 10:04 AM

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.

**yoman360** · September 26th 2009, 11:57 AM

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.

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