can some show me step by step please

thanks

http://img297.imageshack.us/img297/8017/dsc01415.jpg

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- May 18th 2009, 10:07 AMturbod15bmore basic calculus
can some show me step by step please

thanks

http://img297.imageshack.us/img297/8017/dsc01415.jpg - May 18th 2009, 10:15 AMVonNemo19
Show you what. I don' see anything. Are you talking about how you want someone to show you "Step-by-Step" the sitcom? I hated that show.(Angry)

Nevermind, I can see it now. - May 18th 2009, 10:17 AMVonNemo19
f) $\displaystyle \frac{1}{4}x^4+C$ The only step you need for this is the power rule. Do you know the power rule? Oh and algebra too. $\displaystyle dy=x^3dx\Longleftrightarrow\frac{dy}{dx}=x^3$

- May 18th 2009, 10:23 AMVonNemo19
Use the power rule for g too...........

$\displaystyle \frac{1}{3}x^3+x^2-3x+C$

You do know the power rule right? - May 19th 2009, 10:57 AMturbod15b
thanks

but can u show me how to do it for all of them

thanks - May 19th 2009, 11:43 AMdarthfader
j) Let me do j for you step by step. :)

$\displaystyle \int{\frac{{x}}{{x^2 + 2}}}dx$

Let $\displaystyle u = x^2 + 2$

Then $\displaystyle du = 2xdx$

Notice that our $\displaystyle du$ is the same as the numerator of our given except of the numerical coefficient $\displaystyle 2$. Do you see it?

Now, we can put the $\displaystyle 2$ by doing this.

$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}}$ Is this equal to the original given? Of course yes. :) So we can continue.

Then

$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}}$.

There is a formula in integral calculus that states that the $\displaystyle \int{\frac{du}{u}} = \ln\left|u\right| + C$.

Hence in our problem..

$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}} = \frac{1}{2}\ln\left|u\right| + C$.

Plugging in $\displaystyle u$.

$\displaystyle \int{\frac{{x}}{{x^2 + 2}}}dx = \frac{1}{2}\ln\left|x^2 + 2\right| + C $

You can check by getting the derivative of our final answer. If it results to the given, then we're pretty much correct.

I hope this helps.

(Rofl) - May 19th 2009, 11:46 AMturbod15b
thanks bro

can u help me to the other 1s aswell

cheers - May 19th 2009, 12:01 PMdarthfader
For h, let me ask first, is it a Definite Integral? With limits 2 and 3? Do you know how to evaluate Definite Integrals?

The problem is quite easy and you can try it by yourself.

Try to continue this, just feel free to ask if you have some questions.

Hint: Start with this:

$\displaystyle 2\int_2^3 {d\theta + } \frac{1}{2}\int_2^3 {\cos 2\theta (2d\theta )}$

(Rofl) - May 19th 2009, 12:53 PMturbod15b
im lost (Headbang)

- May 19th 2009, 07:24 PMdarthfader
Can you tell me what part of solving the problem you find difficulties?

- May 19th 2009, 07:33 PMVonNemo19
Hey darth! If I were you, I would change my signature. You should write the work

so that it doesn't make you look like you don't know that the limit of a constant is that constant.(Giggle)*function*

I'm just joking with you. Sometimes people can't pick up on humor through text.

The limit as x tends to infinity of work is, guess what?

MORE WORK! Get it?(Rofl) - May 19th 2009, 10:35 PMdarthfader
(Rofl) Thanks. I'll think about changing my signature. Maybe I can come up with a better one. Haha.