# Homework Help: 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

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.