Question

Evaluate the convergent integral.

Answer 1

\[\int\limits_{0}^{9}\frac{ 9 }{ (x-1)^{1/3} }dx\]

Answer 2

Discontinuity at x = 1, \[\lim_{t \rightarrow 1^-}\int\limits_{0}^{t}\frac{ 9 }{ (x-1)^{1/3} }dx+\lim_{t \rightarrow 1^+}\int\limits_{t}^{9}\frac{ 9 }{ (x-1)^{1/3} }dx\]\[\lim_{t \rightarrow 1^-}9\int\limits_{0}^{t}(x-1)^{-1/3}dx=>\lim_{t \rightarrow 1^-}\frac{ 27 }{ 2 }[(x-1)^{2/3}]_{0}^{t}\]\[\lim_{t \rightarrow 1^-}\frac{ 27 }{ 2 }[(t-1)^{2/3}-(-1)]=\frac{ 27 }{ 2 }(0-1)=\frac{ -27 }{ 2 }\]

Answer 3

\[\lim_{t \rightarrow 1^+}\frac{ 27 }{ 2 }[(x-1)^{2/3}]_{t}^{9}=>\lim_{t \rightarrow 1^+}[\frac{ 27 }{ 2 }(8)^{2/3}-0]=\frac{ 27 }{ 2 }(8)\sqrt{8}\]

I'm getting a square root that doesn't simplify cleanly when adding this with (-27/2), the solution is 81/2.

Answer 4

I think that it should be 27/2 rather than -27/2.!

Answer 5

Secondly, the second half is 54 .
Why are you getting it in surd??

Answer 6

I don't see it ...

Answer 7

What didn't you get??
8^2/3 is 4.

Answer 8

Oh I see it, derping myself at this moment. How did I messed up on the first part on how (-27/2) suppose to be positive 27/2 instead? @samigupta8

Answer 9

So you are getting a positive 27/2.
Right??

Answer 10

I got -27/2

Answer 11

Never mind. I got it now, so embarrassing haha. It is that time of the night, I suppose where I'm messing up on the small details.

Answer 12

Okay! How come did you get it ??
You yourself wrote that it was something like
(t-1)^2/3 -(-1)
That means it is (t-1)^2/3 +1
So that simplifies to 1.
Now multiply it by 27/2.

Answer 13

Thanks samigupta8 for the help.

Answer 14

Ah! No problem....