Antiderivative homework question

[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
$$\int_0^h \frac{x}{\sqrt{x}}$$

 

[invisible_man]

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

 

[vorcil]

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

 

[invisible_man]

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

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

 

[vorcil]

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

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

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

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

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

 

[invisible_man]

Oh dear,

x = sqrt(x) * sqrt (x)

 

[vorcil]

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

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

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

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

 

[vorcil]

Oh dear,

x = sqrt(x) * sqrt (x)

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

solved it

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

 

[invisible_man]

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

 

[vorcil]

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 probability 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 probability 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, $$\frac{h}{3}$$

 

[HallsofIvy]

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 probability 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 probability 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, $$\frac{h}{3}$$

Do you mean "voila"?

 

[vorcil]

Do you mean "voila"?

no waaala :P

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

 

[ammontgo]

Vorcil is correct.

 

[Susanne217]

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

 

[Mark44]

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.