So, the integral is: $$\int_1^2\frac{x-2}{\sqrt{x-1}}dx$$ If I copied correctly from the board, the teacher said if x approaches 1+, the function approaches
+$\infty$. What is the difference between 1- and 1+? If x approaches 1+, isn't the function approaching -$\infty$, not +$\infty$.
One more question: When solving that integral, the teacher wrote:
$$\int_1^2\frac{x-1-1}{\sqrt{x-1}}dx=\int_1^2\frac{x-1}{\sqrt{x-1}}dx-\int_1^2\frac{1}{\sqrt{x-1}}dx=$$$$\int_1^2\sqrt{x-1}dx-\int_1^2\frac{1}{\sqrt{x-1}}dx=I1+I2$$
For $I1$, we substitute $x-1$ with $z$ and when we solve, we get $I1=\frac{2}{3}$
For I2, the teacher wrote: $$I2 = \lim_{ɛ->0}\int_{1+ɛ}^2\frac{1}{\sqrt{x-1}}dx$$
And when we substitute $x-1$ with $z$ again, we solve $I2$ and get $I2=1$. My question is, why did the teacher change the integration from $1$ to $2$, to $1+ɛ$ to $2$, what does that $ɛ$ represent?
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A6SE
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Well the integral is improper. So that is why he changed the lower limit from 1 to 1+e as e->0. – randomgirl Mar 28 '15 at 18:01
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Improper because the integrand doesn't exist at x=1. – randomgirl Mar 28 '15 at 18:04
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Oh, I see, thank you very much :) – A6SE Mar 28 '15 at 18:05
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The function $ y=\dfrac{1}{\sqrt{x-1}}$ is not defined for $x=1$, so your second integral is an improper integral that can be evaluated as a limit $$ \lim_{t\rightarrow 1^+}\int_{t}^2\frac{1}{\sqrt{x-1}}dx $$ and the integral exists if this limit exists and is finite.
Writing $1+\epsilon $ for $\epsilon \rightarrow 0$ is the same as $t=1+\epsilon \rightarrow 1$.
Emilio Novati
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The teacher probably wanted to emphasize that the function is increasing from 1 to 2. – A6SE Mar 28 '15 at 18:09
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That's what also bothers me, when we write 1+, why does it differ from simply writing 1? – A6SE Mar 28 '15 at 18:11
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