Evaluating difficult integral involving square roots

hali
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Homework Statement


Evaluate the following integral

Homework Equations


∫ √(4-(√x)) dx


The Attempt at a Solution


I am having a mind block, I find this too challenging, help!
 
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Well, the instant I look at that I think of setting u= 4- \sqrt{x}= 4- x^{1/2}. Then du= -(1/2)x^{-1/2}dx so that dx= -2x^{1/2}du= -2\sqrt{x}du.

And, since u= 4- \sqrt{x}, \sqrt{x}= 4- u.
 
I understand the du = ... But what happens to the square root of the entire function?
 
hali said:
I understand the du = ... But what happens to the square root of the entire function?

That's the whole point of setting ##u = 4 - \sqrt{x}##, the integral then becomes ##\int \sqrt{u} \ (2u-8) \ du ##, you simply substitute u in for ##4 - \sqrt{x}##.

There are other ways of solving this integral as well. You can use trig substituion and you can also set ##u^2 = 4 - \sqrt{x}## and still arrive at the correct result.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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