How to Calculate the Area Under a Curve for v(t) Using Integration?

[brycenrg]

Homework Statement

Show the area under the curve of v(t) is equal to the displacement from t1 to t2

Homework Equations

x/t = v

The Attempt at a Solution

Integrate V(t) = vt dt
(v/2)*t^2]t1 to t2
(v/2)*t1^2 - (v/2)*t2^2

Not sure if that is good enough or how toactually show it. To find the area you take the integration and v(t) is just the derivative of x(t) but I am not how to show it exactly.

 

[Isaac0427]

I would do it like this:
$\int^{t1}_{t2} v(t)dt$
You basically have it.

 

[brycenrg]

Thank you how do we write like that in this forum?

 

[Isaac0427]

Thank you how do we write like that in this forum?

LaTeX. Use two #s to start and end the code.
For more information: https://www.physicsforums.com/threads/introducing-latex-math-typesetting.8997/

 

[Mark44]

Homework Statement

Show the area under the curve of v(t) is equal to the displacement from t1 to t2

Homework Equations

x/t = v

This equation isn't relevant if the velocity isn't constant.

The Attempt at a Solution

Integrate V(t) = vt dt
(v/2)*t^2]t1 to t2
(v/2)*t1^2 - (v/2)*t2^2

Not sure if that is good enough or how toactually show it. To find the area you take the integration and v(t) is just the derivative of x(t) but I am not how to show it exactly.

Since v(t) = $\frac{dx}{dt}$, your integral is $\int_{t_1}^{t_2}v(t) dt = \int_{t_1}^{t_2} \frac{dx}{dt} dt = \int_{t_1}^{t_2} dx$. If you carry that out, what do you get?

 

[brycenrg]

thank you guys. You get t2-t1

 

[haruspex]

thank you guys. You get t2-t1

No. In the final integral in Mark's post, the limit variable and integration variable are different: $\int_{t=t_1}^{t_2}dx$.
What is x when t=t₁?

 

[brycenrg]

Well isn't x = t1 when t is t1
I thought it was x]t2 upper t1 lower
So it's t2 - t1

 

[haruspex]

Well isn't x = t1 when t is t1
I thought it was x]t2 upper t1 lower
So it's t2 - t1

No. x is a position. What is the position at time t1? (so create one!)

 

[brycenrg]

So I could say t1 = 1 and t2 = 2
So then it would be 1 in that case.
So the area would be 1 lol I dono

 

[haruspex]

So I could say t1 = 1 and t2 = 2
So then it would be 1 in that case.
So the area would be 1 lol I dono

No, you can't just plug in arbitrary numbers.
The question asks you to show that the area equals "the displacement from t₁ to t₂". If the displacement x is a function of t, x(t), how would you write the displacement at time t?

 

[brycenrg]

X(t2) - x(t1) is that what they want?

 

[haruspex]

X(t2) - x(t1) is that what they want?

Yes.