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What do the integral ds and dv mean in college physics?
Well, this question tells you an integral formula first.

The integral formula of ghost function ∫ xndx = x (n+1)/(n+1)+c

C is an arbitrary constant, dx is an integral variable, x n is an integrand, and x^ndx is an integrand expression.

I won't give you the deduction, just use it directly.

So we know that the velocity v represents the slope in the s-t diagram, right?

The slope is the tangent of the function image at a point. You should know this, right?

According to mathematical knowledge, we know that when v=lim△t in s-t diagram tends to 0, △ s/△ s/△ t.

For convenience, d is used to represent the trace of high number, so there is no need to write the limit.

Then it becomes v=ds/dt.

Similarly, we can know that a=dv/dt.

This should be known.

Move the above formula to dv=adt.

Bilateral integral ∫dv=∫adt

Think of dv as v^0dv or dv, right? V 00 = 1。

According to the formula I gave you, the left side = v (0+1)/(1+0) = v.

Similarly, the right side is regarded as T 0 to obtain.

That is, v=at

Because v=ds/dt

DS moved to = VDT

It should be v=at, which has been solved above.

Introduced ds = atdt

Similarly, the left side of the formula is exactly equal to S.

A on the right is a constant, so you can get an integer directly without integration.

Then it can be written as s=a∫tdt

Here is the first power of t, and according to the formula, it is s = a * t (1+1)/(1+1) =1/2at2.

At this point, the formula has been deduced, and everyone should understand.

If you want to derive a formula from the initial velocity, add an arbitrary constant c after the integration, which is v0.

After re-integration, S = v0t+ 1/2at 2 can be obtained.