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How do you evaluate the definite integral #int (2x)/sqrt(x^2-1)# from 1 to sqrt5?

Answer 1

Paisley Johnson

The answer is #=4#

We perform this integral by substitution

Let #u=x^2-1#

#du=2xdx#

Therefore,

#int(2xdx)/sqrt(x^2-1)#

#=int(du)/sqrtu#

#=2sqrtu#

So,

#int_1^sqrt5(2xdx)/sqrt(x^2-1)=[2sqrt(x^2-1)]_1^sqrt5#

#=2*2-0#

#=4#

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Answer 2

Isabella Ackerman

To evaluate the definite integral (\int_{1}^{\sqrt{5}} \frac{2x}{\sqrt{x^2 - 1}} , dx), we can start by making a substitution. Let (u = x^2 - 1), then (du = 2x , dx).

This changes our integral to (\int_{0}^{4} \frac{1}{\sqrt{u}} , du), since when (x = 1), (u = 1^2 - 1 = 0), and when (x = \sqrt{5}), (u = (\sqrt{5})^2 - 1 = 4).

Integrating (\frac{1}{\sqrt{u}}) with respect to (u) gives (2\sqrt{u}).

So, (\int_{0}^{4} \frac{1}{\sqrt{u}} , du = 2\sqrt{u} \bigg|_{0}^{4} = 2(\sqrt{4} - \sqrt{0}) = 4).

Therefore, the value of the definite integral (\int_{1}^{\sqrt{5}} \frac{2x}{\sqrt{x^2 - 1}} , dx) is (4).

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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.