Using integration to find limits

waealu
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Homework Statement


The problem asks to integrate dx\(sqrt(abs(x-1))) from 0 to 2.
\int(dx/(sqrt(abs(x-1))))

Homework Equations


I had the limit as b approaches 1 of the integral (from 0 to b) plus the limit as c approaches 1 of the integral (from c to 2).


The Attempt at a Solution


After integrating, I got 2(sqrt(x-1)). Then I plugged in my bounds for the two limit equations and got (0-2)+(2-0)=0. However, I know the answer is 4. What am I missing? I have a feeling that I am forgetting to do something with the absolute value.
 
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waealu said:

Homework Statement


The problem asks to integrate dx\(sqrt(abs(x-1))) from 0 to 2.
\int\frac{}{}dx

Homework Equations


I had the limit as b approaches 1 of the integral (from 0 to b) plus the limit as c approaches 1 of the integral (from c to 2).


The Attempt at a Solution


After integrating, I got 2(sqrt(x-1)). Then I plugged in my bounds for the two limit equations and got (0-2)+(2-0)=0. However, I know the answer is 4. What am I missing? I have a feeling that I am forgetting to do something with the absolute value.

You are forgetting to do something with the absolute value. You can't just integrate pretending it isn't there. |x-1|=x-1 on the interval [1,2] and |x-1|=1-x on the interval [0,1]. Integrate over the two intervals separately and add them.
 
Er Dirk beat me to it.
 
Okay, thanks. I figured it out now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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