# How to use integral

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1. Aug 16, 2015

### Stephanus

Dear PF Forum,
How to use integral?
I created a thread months ago to find out what is the integral of Hubble equation.
Now, I create a thread to understand how to use the integral.

Problem 1:
Something is constantly accelerating 10m/s per meter. Let's say the acceleration is H.
so H = 10m . s-1. h-1
Velocity = H x Distance; V = HD
For a given distance, how can we find the time it takes to cover that distance?
Someone has given me the answer. Okay..., I get it.

Problem 2:
Something is constantly accelerating 10m/s per second. Let's say the acceleration is a
so a = 10 m.s-1.s-1
or a = 10 m/s2
For a given time, how can we find the distance it takes?
I know the answer is D = 1/2 at2. We were taught that at high school.
Instinctively it is the integral of D = VT
I'll try to make a sentence like RyanH42.
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$V = dx/dt$ then $dx = V dt$ then integrate it $V = a.t$ inside the integral will be $\int a.t \, dt$
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Is this sentence correct? I try to understand how to use the integral to solve a problem not to integrate an equation.

2. Aug 16, 2015

### Staff: Mentor

The "at" expression in the integral ∫a.tdt comes from an inner integral ∫adt = at is that what you mean?

3. Aug 16, 2015

### RyanH42

$V=dx/dt$
$dx=Vdt$
$x=∫Vdt$
$a=dV/dt$
$dV=adt$
$V=∫adt$ so

$x=∫Vdt$ here V is equal $V=∫adt$ = $x=∫(∫adt)dt$ a is 10 so, $V=∫adt$, $V=∫10dt$ which its 10t. now $x=∫(∫adt)dt$=$x=∫10tdt$ which its $10t^2/2$

4. Aug 16, 2015

### Staff: Mentor

This doesn't make sense. Acceleration is the time rate of change of velocity, so the units typically are m/s2 or similar (distance/time2).
To solve this problem you have to solve a differential equation, which in this case you can do by integrating twice.
The differential equation is $\frac{d^2s}{dt^2} = a$.
Integrate both sides to get ds/dt, the velocity:
Velocity: $v = \frac{ds}{dt} = \int \frac{d}{dt}\frac{ds}{dt} dt= \int a dt = at + v_0$
Integrate once more to get the displacement:
Displacement $s = \int \frac{ds}{dt} dt= \int v dt = \int (at + v_0)dt = \frac 1 2 at^2 + v_0t + s_0$
The constants v0 and s0 are the initial velocity and initial displacement.

5. Aug 16, 2015

### Stephanus

I mean Distance = $\int a.t \, dt$
But if you said
Is it $V = \int a \, dt$?

6. Aug 16, 2015

### Stephanus

Ahh, this is what @jedishrfu means about inner integral?
So the distance is not straight forward $D = \int a.t \, dt$, but $D = \int (\int a \, dt) \, dt$ while $\int a \, dt = a.t \text { , so } D = \int a.t \, dt$

7. Aug 16, 2015

### Stephanus

Hi Mark44
YES, That's right Mark44, it's NOT acceleration. You might know that the galaxies are traveling from us. The farther, the faster. About 100km/s per 3 million light year. (200 km/s per 6 million ly) You are right! It's not acceleration. But for a non English speaker. I don't know the term for the galaxy movement. Glad you pointed that out.
Yes it makes sense.
It is for the case where the object started not in T0 and at some distance at some velocity. Have to think this paragraph for a while to understand that. It's more complicated than I tought.