# Antiderivative question

• Nov 26th 2009, 01:09 PM
ganur
Antiderivative question
http://img21.imageshack.us/img21/2572/antider.jpg
im pretty sure about everything but the last one..
i guessed it but for some reason i have a good feeling about my guess.

could you please explain it to me?
• Nov 26th 2009, 02:49 PM
hjortur
The last one is correct.

But why is the first one false ? Remember $e$ is just a constant.

$\frac{d}{dx}(x^5+e)=5x^4$

For the second one: if particle has constant acceleration $a$ then
$v(t)=\int adt=at+C_1$

And

$x(t)=\int v(t)=\int at+C_1=\frac{1}{2}at^2+C_1t+C_2$

So it is not a cubic polynomial but a square polynomial.

For the fifth one:

Let F is an antiderivative of f and G is antiderivative of g. Then (using the product rule):

$(FG)'=F'G+FG'$

And that is not fg.
• Nov 26th 2009, 02:52 PM
tonio
Quote:

Originally Posted by ganur
http://img21.imageshack.us/img21/2572/antider.jpg
im pretty sure about everything but the last one..
i guessed it but for some reason i have a good feeling about my guess.

could you please explain it to me?

1 is true (just derivate $x^5 =e$ to check)
2 is false (the position would be a quadratic polynomial)
3 is false
4 is true
5 is false (and big time!)
6 is false

Tonio
• Nov 26th 2009, 02:59 PM
hjortur
hmmm, looks like we disagree on nr 6.

If F and G are antiderivatives of f then $F-G=C$ for some constant $C\in\mathbb{R}$
Now since $F(2)>G(2)$. We know that $F(2)-G(2)=C>0$

Then it follows that $F(4)+C=G(4)$, and since $C>0$,
then $F(4)>G(4)$.

Is this not correct ? If not can you clarify Tonio ?
• Nov 26th 2009, 03:13 PM
tonio
Quote:

Originally Posted by hjortur
hmmm, looks like we disagree on nr 6.

If F and G are antiderivatives of f then $F-G=C$ for some constant $C\in\mathbb{R}$
Now since $F(2)>G(2)$. We know that $F(2)-G(2)=C>0$

Then it follows that $F(4)+C=G(4)$, and since $C>0$,
then $F(4)>G(4)$.

Is this not correct ? If not can you clarify Tonio ?

You are, of course, right. I made a stupid mistake. Thanx

Tonio
• Nov 26th 2009, 03:28 PM
hjortur
Thank you, I was beginning to doubt myself.(Happy)
• Nov 26th 2009, 04:16 PM
ganur
wow thats great help guys! really appreciate that!
though, something isnt working out for me..
for the first one, i had no idea that e is the same as c.. but okay i get it.
2. i get..
5. that makes sense.. yeah i didnt think of that and now im thinking that is so easy (Headbang)

so here are the answer i put in:
T, F, F, T, F, T

and it doesnt work (Wondering)
• Nov 26th 2009, 06:01 PM
hjortur
The trick is that nr 3 is true.

Rewrite $cos^2(x)=\frac{1}{2}+\frac{1}{2}cos(2x)$

Then you get

$2-cos^2(x)=2-\frac{1}{2}-\frac{1}{2}cos(2x)$

And then take the derivative, remember the chain rule.

$\frac{d}{dx}\left(2-\frac{1}{2}-\frac{1}{2}cos(2x)\right)=\frac{1}{2}sin(2x)\cdot2 =sin(2x)$.

I am 100% sure everything other is correct.

Hope that helps.