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How do you integrate #x^(-1/2) - sqrt(x) #?

Answer 1

Samantha Adkison

# 2/3(3-x)x^(1/2)+C.#

We have, #intt^ndt=t^(n+1)/(n+1)+c, n ne -1#.

#:. int{x^(-1/2)-sqrtx}dx#,

#=int{x^(-1/2)-x^(1/2)}dx#,

#=intx^(-1/2)dx-intx^(1/2)dx#,

#=x^(-1/2+1)/(-1/2+1)-x^(1/2+1)/(1/2+1)#,

#=2x^(1/2)-2/3x^(3/2)#,

#=2/3(3-x)x^(1/2)+C.#

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Answer 2

Emily Ammerman

To integrate (x^{-1/2} - \sqrt{x}), you can break it into two separate integrals:

For (x^{-1/2}), integrate using the power rule for integration, which states that ∫(x^n) dx = (\frac{x^{n+1}}{n+1}) + C:

∫(x^{-1/2}) dx = ( \frac{x^{(-1/2)+1}}{(-1/2)+1}) + C = ( \frac{2}{\sqrt{x}}) + C.

For (-\sqrt{x}), use the power rule for integration:

∫(-\sqrt{x}) dx = (-\frac{2}{3}x^{3/2}) + C.

So, the integral of (x^{-1/2} - \sqrt{x}) is ( \frac{2}{\sqrt{x}} - \frac{2}{3}x^{3/2}) + 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.