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?

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- Nov 26th 2009, 01:09 PMganurAntiderivative 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 PMhjortur
The last one is correct.

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

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

For the second one: if particle has constant acceleration $\displaystyle a$ then

$\displaystyle v(t)=\int adt=at+C_1$

And

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

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):

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

And that is not fg. - Nov 26th 2009, 02:52 PMtonio
- Nov 26th 2009, 02:59 PMhjortur
hmmm, looks like we disagree on nr 6.

If F and G are antiderivatives of f then $\displaystyle F-G=C$ for some constant $\displaystyle C\in\mathbb{R}$

Now since $\displaystyle F(2)>G(2)$. We know that $\displaystyle F(2)-G(2)=C>0$

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

then $\displaystyle F(4)>G(4)$.

Is this not correct ? If not can you clarify Tonio ? - Nov 26th 2009, 03:13 PMtonio
- Nov 26th 2009, 03:28 PMhjortur
Thank you, I was beginning to doubt myself.(Happy)

- Nov 26th 2009, 04:16 PMganur
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..

3.&4. i already had that

5. that makes sense.. yeah i didnt think of that and now im thinking that is so easy (Headbang)

6. i understand your argument, and that the answer is 'true'

so here are the answer i put in:

T, F, F, T, F, T

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

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

Then you get

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

And then take the derivative, remember the chain rule.

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

I am 100% sure everything other is correct.

Hope that helps.