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Consider the following integral:

$$I := \int_1^2 \frac{1}{\sqrt{(2-x) (x-1)}} \, dx.$$

The solution of this integral is $I = \pi $.

I typed this integral into Mathematica in a more general form:

Integrate[1/Sqrt[(a - x)*(x - b)], {x, b, a}]

But the output of this is $0$.

Using different variables:

Integrate[1/Sqrt[(x1 - x)*(x - x2)], {x, x2, x1}]

gives the right solution $\pi$.

I don't see the difference between my two inputs. I also used ClearAll[a, b, x1, x2] before evaluating.

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  • $\begingroup$ I am using Wolfram Mathematica 10.2 $\endgroup$
    – drcyberz
    Dec 19 '17 at 12:32
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    $\begingroup$ in ver 11.2 (macOS 10.13.2) ConditionalExpression[\[Pi], a >= b] assuming $a,b\in \mathcal{R}$ $\endgroup$ Dec 19 '17 at 13:09
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Integrate[1/Sqrt[(a - x)*(x - b)], {x, b, a}, Assumptions -> a > b]

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