Integrating the second derivative

**derekjonathon** · November 13th 2009, 07:30 PM

given f''(x) = -25sin(5x) and f'(0) = 3 and f(0) = -2, find f(pi/6)

I am trying to figure out how to integrate the second derivative into the first derivative and make it 3, then integrate into a function of 0 that equals -2.

How do I integrate the second derivative into the first derivative?

**Prove It** · November 13th 2009, 07:37 PM

Originally Posted by derekjonathon

given f''(x) = -25sin(5x) and f'(0) = 3 and f(0) = -2, find f(pi/6)

I am trying to figure out how to integrate the second derivative into the first derivative and make it 3, then integrate into a function of 0 that equals -2.

How do I integrate the second derivative into the first derivative?

$f''(x) = -25\sin{(5x)}$

$f'(x) = \int{-25\sin{(5x)}\,dx}$

$f'(x) = 5\cos{(5x)} + C$ .

You know that $f'(0) = 3$ , so

$3 = 5\cos{(5\cdot 0)} + C$

$3 = 5 + C$

$C = -2$ .

So $f'(x) = 5\cos{(5x)} - 2$

$f(x) = \int{5\cos{(5x)} - 2\,dx}$

$f(x) = \sin{(5x)} - 2x + C$ .

You know that $f(0) = -2$ , so

$-2 = \sin{(5\cdot 0)} - 2\cdot 0 + C$

$C = -2$ .

Therefore

$f(x) = \sin{(5x)} - 2x - 2$ .

Now find $f\left(\frac{\pi}{6}\right)$ .

**derekjonathon** · November 13th 2009, 07:47 PM

It seems easy when you explain it that way!

**redsoxfan325** · November 13th 2009, 07:57 PM

"All truths are easy to understand once they are discovered; the point is to discover them." - Galileo

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Integrating the second derivative

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