derivative/integral problem.

**dandaman** · February 11th 2009, 06:06 PM

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

-my steps
$F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4$
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.

**skeeter** · February 11th 2009, 06:42 PM

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

$F'(2) = 1$

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

$F(2) + 5 = 0$

$F(2) = -5$

so ...

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

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

... you're crazy?

**Chris L T521** · February 11th 2009, 06:43 PM

Originally Posted by dandaman

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

-my steps
$F'(x)=sin(\frac{\pi 2}{4}) - 5=sin(\frac{\pi }{2})-5=1-5=-4$
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, $\frac{\,d}{\,dx}\left[\int_a^xf\!\left(t\right)\,dt\right]=f\!\left(x\right)$

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

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

Does this make sense?

**dandaman** · February 11th 2009, 07:04 PM

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

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

**skeeter** · February 12th 2009, 05:23 AM

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.

**HallsofIvy** · February 12th 2009, 09:15 AM

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

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 $3x^2$ , integrate it to get $x^3+ C$ and then differentiate it, you get $3x^2$ again. But if you do it the other way around, start with $x^3$ , differentiate it to get $3x^2$ and integrate, you get $x^3+ C$ , NOT the same thing.

The same sort of thing happens with functions: $f(x)= x^2$ and $g(x)= \sqrt{x}$ 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 " $\pm$ ".

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