1. ## derivative/integral problem.

$\displaystyle F(x)+5=\int_{2}^{x}sin(\frac{\pi t}{4})dt$
what is $\displaystyle F'(2)+F(2)$

-my steps
$\displaystyle F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4$
then i realized something...
since a derivative and an integral are inverses. Isn't F'(x)+F(x) always gonna equal 0?

I am right or going crazy?

All thanks is appreciated.

2. $\displaystyle F'(x) = \sin\left(\frac{\pi x}{4}\right)$

$\displaystyle F'(2) = 1$

$\displaystyle F(2) + 5 = \int_2^2 \sin\left(\frac{\pi t}{4}\right) \, dt$

$\displaystyle F(2) + 5 = 0$

$\displaystyle F(2) = -5$

so ...

$\displaystyle F'(2) + F(2) = -4$

F'(x) and F(x) are not inverses, and their sum is not always 0.

... you're crazy?

3. Originally Posted by dandaman
$\displaystyle F(x)+5=\int_{2}^{x}sin(\frac{\pi t}{4})dt$
what is $\displaystyle F'(2)+F(2)$

-my steps
$\displaystyle F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4$
then i realized something...
since a derivative and an integral are inverses. Isn't F'(x)+F(x) always gonna equal 0?

I am right or going crazy?

All thanks is appreciated.
You're not quite there.

Recall by the FTC, $\displaystyle \frac{\,d}{\,dx}\left[\int_a^xf\!\left(t\right)\,dt\right]=f\!\left(x\right)$

So in the case of $\displaystyle F\!\left(x\right)+5=\int_2^x\sin\!\left(\tfrac{\pi }{4}t\right)\,dt$, we see that $\displaystyle F'\left(x\right)=\sin\!\left(\tfrac{\pi}{4}x\right )$

Therefore, $\displaystyle F'\!\left(2\right)+F\!\left(2\right)=\sin\!\left(\ tfrac{\pi}{2}\right)+\left[\int_2^2\sin\!\left(\tfrac{\pi}{4}t\right)\,dt-5\right]=1-5=\color{red}\boxed{-4}$

Does this make sense?

4. Yeah it makes sense now... Thanks you two alot.

However, just to be sure, aren't integrals and derivatives inverses of each other? I was 99.9% sure they were.

proof.
$\displaystyle f(x)=x^3$
$\displaystyle f'(x)=3x^2$
$\displaystyle \int 3x^2 dx=x^3 + C$

Am i right now? Or am i still Crazy?lol

5. go see the discussion at this link ...

http://http://www.physicsforums.com/...p/t-84539.html

... hope it will give you some insight. One must use the term "inverse" with extreme care.

6. Originally Posted by dandaman
Yeah it makes sense now... Thanks you two alot.

However, just to be sure, aren't integrals and derivatives inverses of each other? I was 99.9% sure they were.

proof.
$\displaystyle f(x)=x^3$
$\displaystyle f'(x)=3x^2$
$\displaystyle \int 3x^2 dx=x^3 + C$

Am i right now? Or am i still Crazy?lol
Almost. Strictly speaking, two operations, f and g, are inverse if f(g(x))= x and g(f(x))= x. If you start with $\displaystyle 3x^2$, integrate it to get $\displaystyle x^3+ C$ and then differentiate it, you get $\displaystyle 3x^2$ again. But if you do it the other way around, start with $\displaystyle x^3$, differentiate it to get $\displaystyle 3x^2$ and integrate, you get $\displaystyle x^3+ C$, NOT the same thing.

The same sort of thing happens with functions: $\displaystyle f(x)= x^2$ and $\displaystyle g(x)= \sqrt{x}$ are "almost" inverses. If you start with x= 2, square it to get 4 and then take the square root, you get 2 again. But if you start with x= -2, square it to get 4 and then take the squareroot, you get 2, not -2. Of course, to solve a quadratic equation, we take the square root anyway but remember that we need that "$\displaystyle \pm$".