Question

Question #e8b51

Answer 1

One has to do with instantaneous rates of change, while the other mainly deals with the sums of very small components. Or when you have to find the opposite of a derivative.

Explanation:

Covering the obvious ones...
Let's suppose you have a distance-time graph:
graph{0.25x^2 [-6.06, 16.44, -2.43, 8.815]}
And you had to find the velocity at `t=5`.
Since velocity is a rate of change, or the slope, and we have to find the slope of a point `(t, s(t))` we use a derivative. In this case the rate of change is `(ds)/dt`
or the derivative of the function `s` in respect to `t`.

Now, let's suppose you have a question like:
What is the change in velocity of an object in free fall from `0` to `10` seconds since release?

Well, we know that all objects accelerate at `9.8\quadm/s^2` towards the earth. We also know that the derivative of acceleration is velocity.
So, the derivative of what will give us `9.8`?
`d/dt[?]=9.8`
In that case, the what is the integral of `9.8`
`?=int \quad 9.8 \quad dt`
`?=9.8x + C`

But we wanted to find the change in velocity.

So, we take the difference in the velocity from 0 to 10.

`DeltaV=9.8*10-9.8*0=98`

Writing it in shorthand would be:
`DeltaV=int_0^10 9.8 \quad dt`

What's fascinating is that the area under that line 9.8 represents the change in velocity.

Integrals can also be used to find the sum of things divided into very small parts. I'm pretty sure you're familiar with the ol' finding the area under the curve thing.

The sum of all the blue rectangles would be the integral of an interval of the function.

Those are the basics.