Another integral.

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Nov 14th 2010, 08:57 AM
Kristina

Another integral.

Okay,
Can I write like this:
integral x^-4=x^-3/-3 ?
And one more question, from where is 8/9? In this exercise
http://www.part.lt/img/c1e69784c5724...e583256f86.jpg
Nov 14th 2010, 11:54 AM
TheCoffeeMachine

Quote:

Originally Posted by Kristina

Can I write like this:
integral x^-4=x^-3/-3 ?

Yes, $\displaystyle \int{\frac{1}{x^4}\;{dx} = -\frac{1}{3x^3}+k$ .

Quote:

And one more question, from where is 8/9? In this exercise

After differentiating, solve for [LaTeX ERROR: Compile failed] and multiply both sides by $8x^2$ :

$\displaystyle u = 3x^3-1 \Rightarrow \frac{du}{dx} = 9x^2 \Rightarrow dx = \frac{1}{9x^2}\;{du} \Leftrightarrow 8x^2dx = \frac{8}{9}\;{du}.$
Nov 15th 2010, 04:33 AM
Kristina

So,
1)-1/3x^3 fluxcion will be -3x^-3=9x^-4 I'm not right? (Wondering)
2)HOwever I have integration simalar like this example and I tried to use it. But I find problem
Integration (x^2-2) ln(x)dx
u=x^2-2
du=2x
du/dx=2x
dx=du/2x
ln(x)/2x*du
What i do wrong?
Nov 15th 2010, 06:06 AM
harish21

Quote:

Originally Posted by Kristina

So,
1)-1/3x^3 fluxcion will be -3x^-3=9x^-4 I'm not right? (Wondering)

What are you doing here? trying to integrate? $\displaystyle \dfrac{-1}{3x^3} \neq -3x^{_3}$

It is written as $\displaystyle \dfrac{-1}{3x^3} = \dfrac{-x^{-3}}{3}$

Quote:

Originally Posted by Kristina

2)HOwever I have integration simalar like this example and I tried to use it. But I find problem
Integration (x^2-2) ln(x)dx
u=x^2-2
du=2x
du/dx=2x
dx=du/2x
ln(x)/2x*du
What i do wrong?

Is your question $\displaystyle \int (x^2-2)\;ln(x)\;dx$ ???

Then start with $\displaystyle \int (x^2-2)\;ln(x)\;dx = \int x^2\;ln(x)\;dx - \int 2\;ln(x)\;dx$

Use integration by parts for the first term!
Nov 16th 2010, 11:34 AM
Kristina

Okay,
Now I do not understand nothing... (Headbang)
integration u*du dx=(x^3/3-2x*1/x (CAN SOMEONE EXPLAIN HOW TO WRITE FORMULAS?)
Nov 16th 2010, 01:01 PM
mr fantastic

Quote:

Originally Posted by Kristina

Okay,
Now I do not understand nothing... (Headbang)
integration u*du dx=(x^3/3-2x*1/x (CAN SOMEONE EXPLAIN HOW TO WRITE FORMULAS?)

If you posted more clearly and did not confuse your threads with new questions (which I have had to move several times now) you might have a better chance of understanding the help people are giving you.

Click on the link in my signature for how to format equations.

Thread closed.