Beginning Antiderivative Question

[calisoca]

Homework Statement

Find the antiderivative of F(x) \ = \ \sqrt{\frac{5}{x}}

Homework Equations

F(x) \ = \ \sqrt{\frac{5}{x}}

The Attempt at a Solution

1.) F(x) \ = \ \sqrt{\frac{5}{x}}

2.) F(x) \ = \ (\frac{5}{x})^\frac{1}{2}

3.) F(x) \ = \ \frac{5^\frac{1}{2}}{x^\frac{1}{2}}

4.) F(x) \ = \ (5^\frac{1}{2})(x^\frac{-1}{2})

This is where I am stuck. I can find the antiderivative of the x^\frac{-1}{2} term, which would be 2\sqrt{x}, but I'm not sure where to go with the 5^\frac{1}{2} term.

Any help would be greatly appreciated.

 

[Mark44]

You can use the power rule for antidifferentiation. If n != -1,
<br /> \int x^{n} dx<br /> = \frac{x^{n + 1}}{n + 1} + C<br />

 

[jgens]

The sqrt(5) is merely a constant. Integrating functions is similar to differentiating functions. If I asked you to find d(kf(x))/dx where k is a constant, you would likely say kf'(x). Apply similar thinking when applying anti-differentiation.

 

[calisoca]

Well, I know that, for example, if asked to find the antiderivative of 5, the answer would be 5x + C.

I was trying to use similar logic here by saying 5^\frac{1}{2} would become something like \frac{10\sqrt{x^3}}{3}. However, that didn't seem right.

So, if I follow the logic that kf(x) derives to k\frac{df}{dx}, and since 5^\frac{1}{2} is simply a constant, then I would just leave it as is.

Thus, the antiderivative would be (\sqrt{5})(2\sqrt{x}).

Am I understanding this correctly?

 

[Dick]

You are understanding it correctly.

 

[calisoca]

Thank you very much everyone for all of your help!

 

[HallsofIvy]

Well, I know that, for example, if asked to find the antiderivative of 5, the answer would be 5x + C.

I was trying to use similar logic here by saying 5^\frac{1}{2} would become something like \frac{10\sqrt{x^3}}{3}. However, that didn't seem right.

So, if I follow the logic that kf(x) derives to k\frac{df}{dx}, and since 5^\frac{1}{2} is simply a constant, then I would just leave it as is.

Thus, the antiderivative would be (\sqrt{5})(2\sqrt{x}).

Am I understanding this correctly?

Which, of course, is the same as 2\sqrt{5x}+ C so that it is in the same form as the original function (and don't forget the "C").

 

[calisoca]

Yes, of course! Thank you, HallsofIvy.