Question

Can someone explain how this works? My teacher obviously skipped some steps in his explanation :P

Answer 1

Are you doing or starting U-substitution? Basically you're at the end of your calculus 1 unit?

Answer 2

I'm norwegian, so I'm not that familiar with other countries' "level of calculus", but I've had U-subs for half a year now :)

Answer 3

I don't get where the 1 and the 9 comes from! :P

Answer 4

as you put
u =x3+1
now diff wrt x
you get
du/dx=3x^2
3x^2dx=du
put in your equation


now limit

u=X^3+1

previous limit of x are from 0 to 2
put in u
lower limit ;u=0^3+1=1
upper limit; u=2^3+1=9

Answer 5

Aaaah! I guess I wasn't paying enough attention in class that day :P
Thanks alot!

Answer 6

Well, regardless, I'll go out on a limb and say you are.
Your teacher took your original equation, 2(pix) * (3x/(x^3+1)) dx:
\[2pix * \frac{ 3x }{ x^3+1 } dx\]

He factored out the 3, and put the x in the equation, making it

\[6pi * \frac{ x^2 }{ x^3+1 } dx\]

From here, he probably used U-substitution.

So, leave that 6pi out for a second. You're left with x^2 / x^3+1.
He decided that he wanted "U" equal to equal \[x^3+1\]
What's the derivative of that (x^3+1)? It's 3x^2... But since you decided to let some random variable U equal this with terms of x, then you must derive with respect to x! So tag a 'dx' at the end of that.
U = x^3+1
du = 3x^2 dx.

Go back to your original equation.
\[6pi * \frac{ x^2 }{ x^3+1 } dx\]
It looks like your 3x^2 dx can replace a portion of that 6pi, the x^2, and the dx!

So your original equation transformed from:
\[6pi * \frac{ x^2 }{ x^3+1 }dx\]
to
\[2pi * \frac{ 3x^2 }{ x^3+1 }dx\]
Do you see where you can start doing the replacing?
U, from back then, equalled x^3 + 1... Du, from just before, equals 3x^2 dx...
Your new equation:
\[2pi * \frac{ 1 }{ u }du\].

You must have learned that you can factor out a portion of an integral if it's being multiplied.

2pi * the integral of 1/u du.

What equals 1/u du when taking the derivative?
ln|u|, of course. Derivative of ln|u| is 1/u du.

You're left with 2pi times the function going from 0 to 2 of ln|u|.
But you have a U in there! Get rid of it, because it doesn't help us anymore. Remember when U equalled x^3 +1?

Your overall equation now becomes:
\[2pi * function from to 2 of \frac{ ln|x^3+1|\]
Start doing your math:
2pi times the entire quantity of (ln (2^3 + 1) - ln(0^3 + 1)).
2 pi times (ln9 - ln1).

Ln 1 equals 0! Get rid of the ln1.

2 pi times ln9.

Your teacher simplified the ln9 to ln(3^2) because 3 squared gives you nine.
You know that if the equation is 2pi times ln3^2, then the exponent of the ln equation can be carried down.

2 pi times 2 times ln3.

4pi times ln3 is your answer.

Answer 7

I don't know what happened in that little box about half way through. It shouldn't say
"2pi * function from to 2 of \frac{ ln|x^3+1|"

but instead,

"2pi * function from 0 to 2 of ln|x^3+1|"

Answer 8

Thank you so much for taking the time to explain it so thoroughly, hink! :D