If I integrate the speed I obtain the space, but if I integrate the space, what do I get?

Volume?

The integral of speed is distance (one dimension).

11 hours ago, Cristiano said:

If I integrate the speed I obtain the space, but if I integrate the space, what do I get?

When one integrates speed against time, one gets distance (as mathematic already corrects). 

Against what would you like to integrate space? What would the meaning of integrating space against time?

Or try a dimensional analysis. An integral is a sum of products. So it makes sense that if you integrate (m/s) speed over time (s) you get m*s/s = m, so a distance.

Now do this with distance (m) and time, you get m*s. What would that be? OK, via Mr Google I found this: Absement. Had never heard of it...

59 minutes ago, Eise said:

When one integrates speed against time, one gets distance (as mathematic already corrects). 

Against what would you like to integrate space? What would the meaning of integrating space against time?

Or try a dimensional analysis. An integral is a sum of products. So it makes sense that if you integrate (m/s) speed over time (s) you get m*s/s = m, so a distance.

Now do this with distance (m) and time, you get m*s. What would that be? OK, via Mr Google I found this: Absement. Had never heard of it...

 

Wow, you never know when you wake up and log on to SF what new things you might learn.

Thank you for fonding that Eise +1

I have never heard of absement and I couldn't think of a use for the second (or even third fourth etc) integral.

If we take the velocity as some function of t and integrate it we indeed get the distance travelled (but not necessarily the displacement)

To get the second integral we obviously integrate again with respect to t.

I will take the example that v = f(t) = v₁ (a constant) to work from. More complicated functions will necessarily lead to more complicated integrals.

So generally

[math]I = \int {f\left( t \right)dt}  = \int {vdt} [/math]

performing the integration yields

[math]{I_1} = \int {{v_1}dt}  = {v_1}t + C[/math]

performing the second integration yields

[math]{I_2} = \int {\left( {{v_1}t + C} \right)dt}  = \frac{{{{\left( {{v_1}} \right)}^2}}}{2} + Ct + D[/math]

 

 

 

 

Edited July 18, 20187 yr by studiot

6 hours ago, studiot said:

I have never heard of absement and I couldn't think of a use for the second (or even third fourth etc) integral.

Neither did I. Interesting.

While I see possibilities for absement, the higher order integrals (and derivatives beyond jerk ) look like somebody made up a bunch of words and thought it was neat to put them into a Wikipedia article.

53 minutes ago, Bender said:

look like somebody made up a bunch of words and thought it was neat to put them into a Wikipedia article.

Yes I also wondered that.

 

Thank you all.

I've never heard of absement, but unfortunately it doesn't seem useful for my problem, which is related to astronomy. I need to calculate the mean radius vector of an Earth's artificial satellite. But probably this is not the right forum. I first need to ask some basic questions about the ellipse... I switch to the math subforum.