How do you evaluate the integral #int 1/sqrt(x-1)dx# from 5 to #oo#?

Answer 1

#int_(5)^(∞) (1)/(sqrt(x-1)) dx# diverges.

For this particular integral, we can do a u-substitution.

For #int_(5)^(∞) (1)/(sqrt(x-1)) dx#, let #u = x-1 -> du = dx#

Rewriting the integral gives us

#int_(5)^(∞) (1)/(u^(1/2)) du #

Actually, this integral does not converge by the p-series test, which tells us that

#sum_(n=1)^(∞) 1/n^(p)# converges if #p >1# and diverges if #p ≤ 1#.
In our case, since #p = 1/2#, which is surely #≤ 1#, we can conclude that this integral diverges.
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Answer 2

To evaluate the integral (\int_{5}^{\infty} \frac{1}{\sqrt{x - 1}} , dx), we first note that it is an improper integral because it has an infinite limit of integration.

To solve it, we perform a substitution. Let (u = x - 1), then (du = dx).

When (x = 5), (u = 5 - 1 = 4). And as (x) approaches infinity, (u) approaches infinity.

So, the integral becomes (\int_{4}^{\infty} \frac{1}{\sqrt{u}} , du).

This integral converges since (\frac{1}{\sqrt{u}}) approaches zero as (u) approaches infinity. The integral evaluates to (2).

Therefore, (\int_{5}^{\infty} \frac{1}{\sqrt{x - 1}} , dx = 2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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