Beginning Antiderivative Question

1. Jan 19, 2009

calisoca

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Jan 19, 2009

Staff: Mentor

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

3. Jan 19, 2009

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.

4. Jan 19, 2009

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?

5. Jan 19, 2009

Dick

You are understanding it correctly.

6. Jan 19, 2009

calisoca

Thank you very much everyone for all of your help!

7. Jan 19, 2009

HallsofIvy

Staff Emeritus
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").

8. Jan 19, 2009

calisoca

Yes, of course! Thank you, HallsofIvy.