# more basic calculus

• May 18th 2009, 10:07 AM
turbod15b
more basic calculus
can some show me step by step please
thanks

http://img297.imageshack.us/img297/8017/dsc01415.jpg
• May 18th 2009, 10:15 AM
VonNemo19
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 AM
VonNemo19
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 AM
VonNemo19
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 AM
turbod15b
thanks
but can u show me how to do it for all of them
thanks
• May 19th 2009, 11:43 AM
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 AM
turbod15b
thanks bro
can u help me to the other 1s aswell
cheers
• May 19th 2009, 12:01 PM
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.

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

(Rofl)
• May 19th 2009, 12:53 PM
turbod15b
• May 19th 2009, 07:24 PM