Antiderivatives Part I

Hckyplayer8

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Jun 9, 2019
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I took my second exam. No results yet but I will be sure to let you all know once I receive them. No rest for the weary as I'm on the final third of this course.

Moving onto antiderivatives and I just wanted to post this basic problem to make sure I have the concept down.

The indefinite integral of (2x-3) = x2 - 3x + C.

Running it back through differentiation, implies that should be correct.

I'll post some tougher ones later.
 
Thank you. My instructor hasn't covered the rules yet. It's pretty much, its just the reverse of what we have been doing. Reverse engineer based on what you know of the differentiation rules in order to get the answer.
 
Thank you. My instructor hasn't covered the rules yet. It's pretty much, its just the reverse of what we have been doing. Reverse engineer based on what you know of the differentiation rules in order to get the answer.
Derivatives are a bit easier, in a sense, than integration. In many cases it's easy to take a derivative of a function but it can be much harder to take the integral. Most of the basic integration techniques that I know are derived by taking the derivative of some function, say [math]\dfrac{d}{dx} (x^2) = 2x[/math] and then conclude that [math]\int 2x ~ dx = x^2 + C[/math].

-Dan
 
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