# Show the area under a curve

1. Feb 26, 2016

### brycenrg

1. The problem statement, all variables and given/known data
Show the area under the curve of v(t) is equal to the displacement from t1 to t2

2. Relevant equations
x/t = v

3. 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 im not how to show it exactly.

2. Feb 26, 2016

### Isaac0427

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

3. Feb 26, 2016

### brycenrg

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

4. Feb 26, 2016

### Isaac0427

5. Feb 26, 2016

### Staff: Mentor

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

6. Feb 29, 2016

### brycenrg

thank you guys. You get t2-t1

7. Feb 29, 2016

### haruspex

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?

8. Mar 1, 2016

### brycenrg

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

9. Mar 1, 2016

### haruspex

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

10. Mar 1, 2016

### 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

11. Mar 1, 2016

### haruspex

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?

12. Mar 1, 2016

### brycenrg

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

13. Mar 1, 2016

Yes.