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eddysd
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Trying to do an improper integral but for some reason am flummoxed by the integration of 1/sqrt(x-1).
╔(σ_σ)╝ said:Have you tried the substitution [tex]u^{2} = x-1 [/tex]?
Char. Limit said:Why that substitution, I wonder? It seems to me that the substitution u=x+1 would be simpler... but I'm sure that, if I tried yours, it would work out just as easy.
Wait, I did. And it was quite easy. Thanks for the alternative substitution route, I actually like this one better.
Mark44 said:Probably a typo, but u = x - 1 is a better choice than u = x + 1.
The domain of 1/sqrt(x-1) is x > 1, since the function is undefined for x = 1. The range is all real numbers greater than or equal to 0, since the square root of any number can never be negative.
To solve the integral of 1/sqrt(x-1), you can use the substitution method. Let u = sqrt(x-1), then du/dx = 1/(2*sqrt(x-1)). This means that dx = 2*sqrt(x-1)*du, which you can then substitute into the integral. This will result in the integral becoming the more manageable form of ∫1/u du, which can be easily solved using the power rule.
Yes, there is a shortcut called the trigonometric substitution method. You can let x = sec^2(theta), which will result in dx = 2tan(theta)*sec(theta)d(theta). This substitution will transform the integral into ∫1/tan(theta)sec(theta)d(theta), which can be solved using the double angle formula for tangent.
The integral of 1/sqrt(x-1) can be used in real-world applications involving rates of change. For example, it can be used to calculate the velocity of an object moving along a curved path with a varying radius of curvature, or to calculate the rate of change of temperature in a system with a varying heat transfer coefficient.
Yes, the graph of 1/sqrt(x-1) can be easily plotted using various online graphing tools or software. It will show a curve that approaches infinity as x approaches 1 from the right, and gradually decreases as x increases. The graph will also show a vertical asymptote at x = 1.