Having a problem in steps while solving integrals

[prakhargupta3301]

Homework Statement

My problem is in integral calculus (I'm new to it).
I know what it is and how it works (basically. I'm not too advanced right now).
The problem is as following: (I will be posting comments/reasons along with what I've done and with what logic/understanding I've done it. Please correct me if I'm wrong.)
I was trying to derive kinematic equations using differentiation.
a=d(v)/dt -{1. here, we differentiated velocity w.r.t time}
∫adt= ∫d(v) -{2. here, we have multiplied both sides by dt. I don't know how that's even possible but my teacher told it that way. If there is some other underlying logic, please be kind enough to enlighten me.}
at|^v_u= v -{3. a is taken as a constant. So, it comes out of the integration as it is. Integral of dv is v( how? I don't know. My teacher told it this way. (He explained it this way: Sum of multiple small dv's will be v. I got it during the class, but now I have a doubt. With respect to what are integrating dv?)
at=v-u
Derived.

Homework Equations

x_x_x

The Attempt at a Solution

This isn't a problem but a doubt in my understanding. So there are no attempts. I have already written what all there is.
Please help.

 

[PeroK]

Those are standard steps so there isn't a lot to say except that's the way integration works. I would, however, do it differently. Starting with

$a = \frac{dv}{dt}$

I would integrate that equation with respect to $t$.

$\int_{t_0}^{t_1} a dt = \int_{t_0}^{t_1} \frac{dv}{dt} dt$

Now, remembering that the integral is an anti derivative, we have

$\int_{t_0}^{t_1} a dt = v(t_1) - v(t_0)$

And, if $a$ is a constant, we have:

$a(t_1 - t_0) = v(t_1) - v (t_0)$

 

[BvU]

Hi,

Perhaps it helps you if you rephrase the $d$ in $dt$ by

" $\Delta$ with the understanding that we take the limit $\Delta \downarrow 0$ "

that way you 'see' that $a \,dt = dv$ and all of those manipulations that physicists do without much consideration. For mathematicians it can be a horror to look at them, but in physics functions are generally well-behaved enough to talk like this.

 

[prakhargupta3301]

I would integrate that equation with respect to ttt.

∫t1t0adt=∫t1t0dvdtdt∫t0t1adt=∫t0t1dvdtdt\int_{t_0}^{t_1} a dt = \int_{t_0}^{t_1} \frac{dv}{dt} dt

Now, remembering that the integral is an anti derivative, we have

∫t1t0adt=v(t1)−v(t0)∫t0t1adt=v(t1)−v(t0)\int_{t_0}^{t_1} a dt = v(t_1) - v(t_0)

So, if there is a differentiation inside an integral, then what happens? I don't get how you just take it (v) out. (I understand differentiation is just opposite of integration, but I'm new to one being inside the other. Can you break it a bit more please?)
(I don't get what is wrong with the reply mech. Please don't mind the raw code for the equation. I don't know how this works so please forgive me)
Thank you.

 

[PeroK]

So, if there is a differentiation inside an integral, then what happens? I don't get how you just take it (v) out. (I understand differentiation is just opposite of integration, but I'm new to one being inside the other. Can you break it a bit more please?)
(I don't get what is wrong with the reply mech. Please don't mind the raw code for the equation. I don't know how this works so please forgive me)
Thank you.

That the integral of the derivative of a function is the function is the Fundamental Theorem of Calculus.

You need to know that before you can move on with integration.

 

[prakhargupta3301]

Hi,

Perhaps it helps you if you rephrase the $d$ in $dt$ by

" $\Delta$ with the understanding that we take the limit $\Delta \downarrow 0$ "

that way you 'see' that $a \,dt = dv$ and all of those manipulations that physicists do without much consideration. For mathematicians it can be a horror to look at them, but in physics functions are generally well-behaved enough to talk like this.

So, you're trying to say that mathematically, this is wrong, but in physics such manipulations work so I needn't worry?
Also, Just to be sure, what you wrote in the double quotation marks means: Change where we take the lowest limit to be nearly equal to 0?
Thank you.

 

[prakhargupta3301]

That the integral of the derivative of a function is the function is the Fundamental Theorem of Calculus.

You need to know that before you can move on with integration.

Ok. I think I get it now. (Perhaps I'm getting these many silly doubts is because we aren't being taught calculus in totality but bits and pieces of it in Mathematical Tools section of physics class.)
Also, earlier I was thinking of v as a constant (I know that was ignorant of me) but I now realize it's a function!
Anyways, I checked it and now I know the Fundamental theorem of Calculus. Thanks to you!