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Some help with Calculus 2

Integrate 1/(x-1)^(2/3) dx with bounds (0,2).

August 7, 2013

6 Comments • Newest first

CaptCandy

[quote=CursedVoyboy]Times x by 2/3 equals 1.

Lol s0 dumb[/quote]

Can't tell if seriously being dumb or not, but OT said integrate (area under the curve, ahem)

@lollol656: First, we must check if it converges, as x to 1 from the left, it approaches -infinty, and as it approaches from right, it goes to infinity so it converges (we're checking for discontinuities), so at 1, it becomes impossible to integrate... In this state.

Regarding @hyperfire7's advice, you can break it into two and do limits as it approaches 1, or, an easier method in this case would be the substitution method.

So instead of integrating (x-1)^-(2/3) from 0 to 2, substitute with u = x - 1,
and it becomes integrate (u)^-(2/3) from u = -1 to 1 (because the limits of integration also are affected by the substitution)

From here, the antiderivative is going to be 3 * u^(1/3), and from -1 to 1, so that would be 3 * (1)^(1/3) - 3 * (-1)^(1/3) = 3 + 3 = 6.

The answer is 6, whether you use this technique or @hyperfire7's technique.

Reply August 7, 2013 - edited
hyperfire7

[quote=lollol656]so the anti-derivative is 3(x-1)^(1/3). how do I know if it converged?[/quote]
Solve it all the way through. If either integrals give an undefined answer, it diverges.
I solved it, the answer should be 6.

Reply August 7, 2013 - edited
lollol656

so the anti-derivative is 3(x-1)^(1/3). how do I know if it converged?

Reply August 7, 2013 - edited
hyperfire7

[quote=OhShutUp][url=http://www.wolframalpha.com/input/?i=integral+%280%2C2%29+1%2F%28x-1%29%5E%282%2F3%29+dx]it's been like two years since i've been in calc but idk try this[/url]
lmao i don't remember a thing sorry if that is no help[/quote]
Shouldn't get an imaginary answer b/c the integral converges.

Reply August 7, 2013 - edited
OhShutUp

[url=http://www.wolframalpha.com/input/?i=integral+%280%2C2%29+1%2F%28x-1%29%5E%282%2F3%29+dx]it's been like two years since i've been in calc but idk try this[/url]
lmao i don't remember a thing sorry if that is no help

Reply August 7, 2013 - edited
hyperfire7

Alright so you can't integrate this normally since when x=1, the integration will give an undefined answer.
Break it apart using limits, and let's call the function f(x):
(lim a->1- of integral f(x) with bounds 0 and a) + (lim b->1+ of integral f(x) with bounds b and 2).
Everything else should be stuff from Calc I.

Reply August 7, 2013 - edited