What is the general antiderivative of this complicated function?

AI Thread Summary
The discussion centers on finding the general antiderivative of a complex function, specifically f'(x) = 1 - 2x - 4/√x + 5/x - 8/(1+x²) + 9/x⁴. Participants clarify that the goal is to identify the original function f(x) from its derivative f'(x). They emphasize the importance of separating each term to integrate them individually and remind that a constant of integration, +C, must be included in the final answer. The conversation highlights the challenges faced due to missed class time and the need for a solid understanding of antiderivatives. Overall, the thread provides guidance on approaching the problem and reinforces key concepts in calculus.
frumdogg
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Homework Statement



Hi everyone. My Calc 1 final exam is tomorrow and due to some weather related issues we were not able to cover all material for this quarter. With that said, my professor gave us a take home quiz on material that was unable to be covered. I have done by best, but I am getting hung up on an antiderivative problem.

Find the general antiderivative of:


f'(x) = 1-2x-4/\sqrt{x}+5/x-8/(1+x^2)+9/x^4

Homework Equations



Now due to having virtually no time to learn about antiderivatives (we lost a whole week due to a blizzard and instructor illness) I am really unsure where to go. Do I need to rewrite the problem on one line and then find the opposite of the derivative?

Thanks!
 
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well yes, what function, if you differentiate it, will give that expression as its derivative?
 
So basically the definition of an antiderivative is the oppositive of the derivative?

Wait, I did not just type that last sentence. =)
 
Do you mean:

f'(x) = 1-2x-4\sqrt{x}+\frac{5}{x}-\frac{8}{1+x^2}+\frac{9}{x^4} ?

Basically when you're asked to find the anti-derivative you're trying to find the function f(x), which has this derivative f'(x), which is given.
 
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1- 2x -4\sqrt{x}+\frac{5}{x}-\frac{8}{1+x^{2}}+\frac{9}{x^{4}}
 
yes!

But you often write it given f(x), find its primitive function F(x): F' = f
 
That is sort of how I figured that it would work. My derivative skills are giving me the most trouble at this point.
 
Do you have a textbook? First, read the section on anti-derivatives.

Hint: separate each quantity
f '(x) = 1, then f(x) = x + C.
f '(x) = 2x, then f(x) = ?
f '(x) = 9x^-4, then f(x) = ?
.
.
.
and so on.
 
f '(x) = 2x, then f(x) = x^2
f '(x) = 9x^-4, then f(x) = 9x^-3/-3
correct?
 
  • #10
The general antiderivative will have a + C at the end.
 
  • #11
frumdogg said:
f '(x) = 2x, then f(x) = x^2
f '(x) = 9x^-4, then f(x) = 9x^-3/-3
correct?

So far, so good

But Snazzy is right you need to add a +C (constant of integration) term to the anti derivative, because if f(x) had a constant value some where in it, like f(x) = 2x+5, f'(x) = 2 -- so when we integrate f'(x) we need to account for the 5. We don't necessairly know it's a 5 so that's why we add the +C
 
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  • #12
It's so easy to forget the C at the end.
 
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