# Antiderivative homework question

1. Apr 14, 2010

### vorcil

Basically I need to know $$\int_0^h x\frac{1}{2\sqrt{hx}}dx$$

my working,

$$\frac{1}{2\sqrt{hx}} \int_0^h \frac{x}{\sqrt{x}}$$

,

how do i do this?,
$$\int \frac{1}{\sqrt{x}}$$

$$\int x^(\frac{-1}{2})$$

$$\frac{1}{\frac{-1}{2}+1} x^(\frac{-1}{2}+1)$$

$$\frac{x^\frac{1/2}}{\frac{1}{2}}$$

$$2x^\frac{1}{2}$$

but do i just multiply that by the anti derivative of x? since i want
$$\int_0^h \frac{x}{\sqrt{x}}$$

2. Apr 14, 2010

### invisible_man

Re: Antiderivative

you got wrong in "my working" . x is variable. so you can't take it out of the integral. 2sqrt(h) is constant so you can take to the front.

1/2sqrt(h) * integral sqrt(x) dx from 0 to h.

= 1/2sqrt(h) * 2/3 * x^3/2 and evaluate at x = h.

= 1/2sqrt(h) * 2/3 * h sqrt(h) = h/3

3. Apr 14, 2010

### vorcil

Re: Antiderivative

I didn't take x out of the integral

all i did was integrate $$\frac{1}{\sqrt{x}}$$

How did you get to $$\frac{2}{3} x^\frac{2}{3}$$???

-

i want $$\int \frac{x}{\sqrt{x}}$$

I don't know how

4. Apr 14, 2010

### invisible_man

Re: Antiderivative

you have the integral of x / sqrt(x) = integral of sqrt(x).

integral of sqrt(x) = 2/3 * x^3/2

5. Apr 14, 2010

### vorcil

Re: Antiderivative

$$\frac{x}{\sqrt{x}} how does that become \sqrt{x} ?$$

$$x*x^\frac{-1}{2} = x^(1+(\frac{-1}{2})) = x^\frac{1}{2}$$

$$\int x^\frac{1}{2} = \frac{x^\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{3}{2}x^\frac{2}{3}$$

$$\int_0^h \frac{2}{3} x^\frac{2}{3} = \frac{2}{3}h^\frac{2}{3}$$

Last edited: Apr 14, 2010
6. Apr 14, 2010

### invisible_man

Re: Antiderivative

Oh dear,

x = sqrt(x) * sqrt (x)

7. Apr 14, 2010

### vorcil

Re: Antiderivative

$$\frac{1}{2\sqrt{h}}\frac{2}{3}h^\frac{2}{3}$$

crossing out 2s,

$$\frac{1}{\sqrt{h}}\frac{1}{3}h^\frac{2}{3}$$

realizing h^2/3 is the same as h^1 * h^0.5

$$\frac{h^1*h^\frac{1}{2}}{h^\frac{1}{2}3}$$

$$\frac{h}{3}$$

8. Apr 14, 2010

### vorcil

Re: Antiderivative

that dosen't help me at solving $$\frac{x}{\sqrt{x}}$$
at all...

solved it

$$x*x^\frac{-1}{2} = x^(1+(\frac{-1}{2})) = x^\frac{1}{2}$$

9. Apr 14, 2010

### invisible_man

Re: Antiderivative

what is your question? take that integral or what? type it out your problem and question so I can understand it

10. Apr 14, 2010

### vorcil

Re: Antiderivative

I solved it but,

the initial question was:

---------------------------------------------------------------------------------
Suppose i drop a rock off a cliff of height h, as it falls i snap a million photographs at random intervals, on each picture i measure the distance the rock has fallen.

question: what is the average of all these distances? that is to say, what is the time average of the distance travelled?

-------------------------------
what i did,

The rock starts out at rest, and picks up speed as it falls,

x(t) = 1/2 gt^2

velocity = dx/dt =gt, since v=at and a=g,
the total flight time is $$T=\sqrt{\frac{2h}{g}$$

the probability the camerea flashes in the interval dt, is dt/T,
so the probablity that a given photograph shows a distance in the coressponding range dx is,

$$\frac{dt}{T} = \frac{dx}{gt} \sqrt{\frac{g}{2h}} = \frac{1}{2 \sqrt{hx}}dx$$

the probablity density that (0<=x<=h),

$$p(x) = \frac{1}{2\sqrt{hx}}$$

so that the rock can't be bellow the ground or above from where it was dropped,

putting that into the equation,

$$1=\int_{-\infty}^\infty p(x)dx$$

i get,

$$\int_0^h \frac{1}{2\sqrt{hx}}dx = \left[\frac{1}{2\sqrt{h}}(2x^{\frac{1}{2}}]\right|_0^h =1$$

and answering the question, what is the average distance,

<X>

i get by putting it into the equation,

$$<x> = \int_{-\infty}^\infty xp(x)dx$$

and this is where i needed help with the maths, just for the little algebra/calculus bits,

$$<x> = \int_0^h x \frac{1}{2\sqrt{hx}}dx = \frac{1}{2\sqrt{h}} \int \frac{x}{\sqrt{x}} = \left[ \frac{1}{2\sqrt{h}} (\frac{2}{3}x^\frac{3}{2}) \right|_0^h$$

and wala, $$\frac{h}{3}$$

Last edited: Apr 14, 2010
11. Apr 15, 2010

### HallsofIvy

Re: Antiderivative

Do you mean "voila"?

12. Apr 15, 2010

### vorcil

Re: Antiderivative

no waaala :P

i actually calculated the standard deviation of this also :P wondering if i should post it here

13. Apr 19, 2010

### ammontgo

Re: Antiderivative

Vorcil is correct.

14. Apr 21, 2010

### Susanne217

Re: Antiderivative

Why do it so complicated this is posted in the pre-calculus forum.

if $$\int_{0}^{h} \frac{x}{2 \sqrt{hx}}dx = \int_{0}^{h} \frac{x}{2} \cdot \frac{1}{\sqrt{hx}} dx$$

where by using integration by parts I get

$$u = \frac{x}{2} \Rightarrow du = \frac{dx}{2}$$ and $$dv = \frac{dx}{\sqrt{hx}} \Rightarrow v = \frac{2 \sqrt{hx}}{h}$$

Thus

$$\bigg[ \frac{x \cdot \sqrt{hx}}{h}\bigg]_{0}^{h} - \int_{0}^{h} \frac{\sqrt{hx}}{h} dx = h - \frac{1}{h} \int_{0}^{h} \sqrt{hx} dx = h - \frac{1}{h} \cdot \bigg[\frac{2\cdot x \cdot \sqrt{hx}}{3}\bigg]_{0}^{h} = h - \frac{1}{h} (\frac{2h^2}{3}) = \frac{h}{3}$$

Susanne

Last edited: Apr 21, 2010
15. Apr 21, 2010

### Staff: Mentor

Re: Antiderivative

Although this integral can be calculated using integration by parts, why would you want to use this technique when a much simpler technique is available? The integrand can be written as a constant times x^(1/2), so the antiderivative will be the constant times (2/3) x^(3/2).

No one else seems to have noticed (including myself) that the integral is actually an improper definite integral, due to the integrand being undefined at the lower integration limit. The value of the integral is unaffected, but it should be taken into account.