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How do you integrate #dy / (4(y^(1/2))#?

Answer 1

Emilia Aguilar

#1/2y^(1/2)+C#

#[1]" "intdy/(4(y^(1/2)))#

First you can bring #1/4# outside the integral symbol.

#[2]" "=1/4intdy/y^(1/2)#

Next, bring #y^(1/2)# to the numerator.

#[3]" "=1/4inty^(-1/2)dy#

Use the power rule: #intx^ndx=x^(n+1)/(n+1)+C# (where C is a constant)

#[4]" "=1/4*y^(-1/2+1)/(-1/2+1)+C#

#[5]" "=1/4*y^(1/2)/(1/2)+C#

#[6]" "=color(red)(1/2y^(1/2)+C)#

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Answer 2

Charles Arocha

To integrate dy / (4(y^(1/2))), you can use the substitution method. Let u = y^(1/2), then dy = 2u dy. Substituting these into the integral, you get: ∫(dy / (4(y^(1/2)))) = ∫(1 / (4u)) * (2u dy) = (1/2) ∫du = (1/2) * ln|u| + C = (1/2) * ln|y^(1/2)| + C = (1/2) * ln|√y| + C.

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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.