# Relationship between Differentiation and Integration?

1. Jun 27, 2012

### Concavity

Hello everyone, I have a question that I have spent many nights pondering and hours on my whiteboard considering. I apologize in advance if this question seems a bit elementary, but to me it is something that I believe is all important before I can understand all of calculus.

How is differentiation, the use of limits that allows us to find the instantaneous rate of change (the slope), interals of increase/decrease, and one of the ways of deriving functins of velocity and acceleration

the opposite of

Integration, definite or indefinite, which is usually described as the limit of a infinite sum of areas under a curve(definite), or a family of functions (indefinite)?

I understand the mathematical justifications and proofs involving anti differentiation and the fundamental theorems of calculus, what I struggle with Is the Conceptual connection. My math teacher last year struggled to explain it to me but he said it had to do with D=vt?

Anyone's help is much appreciated:)

2. Jun 27, 2012

### Vorde

Well to the specific question of the relationship of velocity to distance.
Can you understand that Velocity is the rate at which distance changes? If so then the derivative (which is defined as the rate of change) of position would be velocity (and the same to acceleration).

As to the more broad integral/derivative question.
I struggled with the exact same thing for a while. What got me over it was this. Define integration as the opposite of differentiation, don't worry about the implications for area yet and just know that integration undoes differentiation. Then consider a function and its derivative. The derivative is the amount by which the function is changing, so by using the concept of differentials, a little bit of the derivative corresponds to the change in y of the original function, and as little bit becomes infinitely small, the change in y converges to absolute accuracy. Then realize that the sum of all of the little changes in y can be represented (without having any idea of the inbetween changes) by the ending value of the function minus the beginning value.

Combining this: the change in a function is equal to the sum of all the little changes of a function. Then realize that 'all the little changes' is equivalent to the derivative of the function times the differential dx. So you can write that:
∫ $\frac{dy}{dx}$ dx = total change

Then because the sum of changes is equal to the end value minus the beginning, you get the fundamental theorem of calculus.

Last edited: Jun 28, 2012
3. Jun 28, 2012

### SteveL27

Yes, the Fundamental Theorem of Calculus is just distance = rate * time on steroids.

Think of it this way. I'm in my car. I drive about 50 mph for a few minutes; then I drive about 45 mph for a few minutes; then I drive about 60 mph for a few minutes. To estimate how far I went, I take

(50mph * number of minutes at 50) + (45mph * number of minutes at 45) +
(60mph * number of minutes at 60). That's a decent approximation. It's approximate because within each interval my speed wasn't constant, but we just called it constant so we could use d = rt.

Now if you do the same thing over intervals of one second each instead of a few minutes ... you get a more accurate estimate. If you let the interval go to zero you get the FTC.

Last edited: Jun 28, 2012
4. Jun 28, 2012

### HallsofIvy

Another way of thinking about it (I get the feeling you want a general, intuitive, grasp rather than specific formulas) is that the derivative is "division" (it generalizes "slope" of a line which is "rise/run") and integration is multiplication (it generalizes "area" of rectangle which is "length times height"). That is the basic reason they are inverse operations.

Last edited by a moderator: Jun 28, 2012
5. Jun 28, 2012

### Concavity

Wow. I am truly thankful for all of your answers, they all allowed me to look at my question in different ways and it kind of just clicked for me. So thank you 3 very much! It really helped me out,
- Concavity

Also, when I was reading these it occurred to me that perhaps I can prove this graphically with time as the independent variable on the x axis and d(t) and v(t) plotted on the axes, observing their relations visually.

6. Jun 28, 2012

### Vorde

Yep.

I don't know if you have taken physics yet, but I took physics before I took calculus and looking back so much makes perfect sense when I look at things in terms of derivatives and integrals.