1. ## General Antiderivatives

Just got this question wrong, I make a mistake somewhere and I need some pointing out where to where it is. Heres the problem first

Find the most general antiderivative of the function
Alright.. Here's my work:

First I converted the cube root x^2 and sqrt x^9 to :

f(x) = x^2/3 - x^9/2

Then using the antiderivative rule of $\displaystyle x^(n+1)/n+1$ , I got:

f(x) = (x^5/3) / (5/3) - (x^11/2) / (11/2) + C

simplifying that I get :

3/5x^(5/3) - 2/11x^(11/2) + C

... I ended up getting the answer wrong

Help is greatly appreciated

2. Originally Posted by endiv
Just got this question wrong, I make a mistake somewhere and I need some pointing out where to where it is. Heres the problem first

Alright.. Here's my work:

First I converted the cube root x^2 and sqrt x^9 to :

f(x) = x^2/3 - x^9/2

Then using the antiderivative rule of $\displaystyle x^(n+1)/n+1$ , I got:

f(x) = (x^5/3) / (5/3) - (x^11/2) / (11/2) + C

simplifying that I get :

3/5x^(5/3) - 2/11x^(11/2) + C

... I ended up getting the answer wrong

Help is greatly appreciated
your answer is correct. the integral is $\displaystyle \frac 35x^{\frac 53} - \frac 2{11}x^{\frac {11}2} + C$

3. $\displaystyle \begin{gathered} f'(x) = \sqrt[3]{{x^2 }} - \sqrt {x^9 } = x^{\frac{2} {3}} - x^{\frac{9} {2}} \hfill \\ f(x) = \frac{3} {5}x^{\frac{5} {3}} - \frac{2} {{11}}x^{\frac{{11}} {2}} + C \hfill \\ \end{gathered}$

4. Oh, wow. My mistake was putting a little c instead of a capital C, which resulted my in a wrong answer :|. Thanks!

5. Originally Posted by endiv
Oh, wow. My mistake was putting a little c instead of a capital C, which resulted my in a wrong answer :|. Thanks!
i don't see why that should make a difference, but ok. maybe whatever you are typing the answer into is case sensitive