What is the general antiderivative of this complicated function?

[frumdogg]

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!

 

[malawi_glenn]

well yes, what function, if you differentiate it, will give that expression as its derivative?

 

[frumdogg]

So basically the definition of an antiderivative is the oppositive of the derivative?

Wait, I did not just type that last sentence. =)

 

[Feldoh]

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.

 

[frumdogg]

1- 2x -4$$\sqrt{x}$$+$$\frac{5}{x}$$-$$\frac{8}{1+x^{2}}$$+$$\frac{9}{x^{4}}$$

 

[malawi_glenn]

yes!

But you often write it given f(x), find its primitive function F(x): F' = f

 

[frumdogg]

That is sort of how I figured that it would work. My derivative skills are giving me the most trouble at this point.

 

[scarecrow]

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.

 

[frumdogg]

f '(x) = 2x, then f(x) = x^2
f '(x) = 9x^-4, then f(x) = 9x^-3/-3
correct?

 

[Snazzy]

The general antiderivative will have a + C at the end.

 

[Feldoh]

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

 

[frumdogg]

It's so easy to forget the C at the end.