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How do you evaluate #int (1) / (sqrt(1 + x))# for [0, 3]?

Answer 1

Evelyn Agard

You can rewrite this as:

#= int (1+x)^"-1/2"dx#

and now you can simply use the reverse power rule.

#= [2(1+x)^"1/2"]##|_(0)^(3)#

#= [2sqrt(1+x)]##|_(0)^(3)#

#= [2sqrt(1+3)] - [2sqrt(1+0)]#

#= 4 - 2#

#= color(blue)(2)#

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Answer 2

Cameron Alexander

To evaluate the integral (\int_0^3 \frac{1}{\sqrt{1 + x}} , dx), you can use the substitution method. Let (u = 1 + x), then (du = dx). When (x = 0), (u = 1), and when (x = 3), (u = 4). Substituting these values, the integral becomes (\int_1^4 \frac{1}{\sqrt{u}} , du), which equals (2\sqrt{u}). Evaluating this from 1 to 4 gives (2(\sqrt{4} - \sqrt{1}) = 2(2 - 1) = 2). Therefore, the value of the integral is 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.