How come the inverse of a derivative is an integral?

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I can’t conceptually grasp this the same why I can grasp addition/subtraction (the opposite of adding stuff is to take it away), but if the derivative of a function is the slope at any given point, then why would the area under the curve be the opposite of that?

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Anonymous 0 Comments

Division is the opposite of multiplication.

Differentiation involves taking a thing, finding a tiny change in how far up it goes (B), a tiny change in how far along it goes (A), and dividing them to get the derivative (C = B/A).

Integration involves taking a thing (C), multiplying it by a tiny change in along (A), and getting the “area” or integral (B = A * C).

Ignoring the fancy infinitesimals and finite sums stuff, we have the same equation, just one using division, one using multiplication.

We can go more into this; if we increase the upper limit on an integral, we’re adding on an extra slice – that slice will have a “height” of the value of the function at that point. So the rate of change of our integral is just the function.

For differentiation, if we try to find the area under our rate of change curve, it will give us an infinite sum of the infinitely small changes in our function as we move along. Well if we add together how much our function has gone up by from some starting point to some end point, it will give us the actual value at that end point (up to a constant).

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