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Sherlock
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- Jan 28, 2012
- 59
Consider $\displaystyle I = \displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}.$ It can be shown that this is $\pi$.
Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$
However, the latter integral is equal to $-\pi$. Where is the negative from?
Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$
However, the latter integral is equal to $-\pi$. Where is the negative from?
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