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Antidifferentiation - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
The anti-derivative of 0 is a constant. We know this because the derivative of a constant is always 0.
The anti-derivative of a constant m is some function of x, mx + c, where c is another constant. This is because the derivative of mx + c is m. The constant c is important to remember -- there are infinite possibilities for what this c might be, but it becomes 0 when you take the derivative.
Let’s do a problem where we find some anti-derivatives. And let’s start really simple. I want to find all the anti-derivatives of first f(X) equals 0. And then f(x) equals m, some constant. Starting with 0. What does it mean to find a function whose anti-derivatives. Find the anti-derivatives of little f(x) equals 0.
We are looking for functions whose derivative is 0. Those functions are the constant functions. At every point their slope is horizontal so the derivative is 0. And a function like this would be just some constant function. So f(x), capital F(x) equals c. That will be the set of all anti-derivatives of little f(x) equals 0. If you differentiate this you get 0. And that proves that these are the anti-derivatives of 0. So f(x) equals c are the anti-derivatives of 0, of little f(x) equals 0. And I always try to distinguish my anti-derivatives with a capital letter and my original function with a lower case.
So let’s take a look at anti-derivatives of little f(x) equals m. First of all, what does that mean? It means functions whose derivative is m. So I can think of one right off the top of my head. Capital F(x) equals mx. Its derivative is m take the derivative you get m. And so that does work as one anti-derivative.
And remember we had a theorem that says, if you find one anti-derivative you can find all of them just by adding a constant to this. So f(x) equals mx plus c. I can add any constant to this mx and I’ll get an anti-derivative of m because its derivative would still be m. The derivative of the plus c part is just 0. So this constitutes the set of all the anti-derivatives of m. So the anti-derivatives of little f(x) equals m are F(x) equals mx plus c.
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