# General Antiderivatives

• Nov 16th 2008, 03:44 PM
endiv
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

Quote:

Find the most general antiderivative of the function
http://i137.photobucket.com/albums/q...dc3aaf8e28.gif
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 $x^(n+1)/n+1$ , 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 (Bow)
• Nov 16th 2008, 03:55 PM
Jhevon
Quote:

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 $x^(n+1)/n+1$ , 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 (Bow)

your answer is correct. the integral is $\frac 35x^{\frac 53} - \frac 2{11}x^{\frac {11}2} + C$
• Nov 16th 2008, 03:57 PM
Plato
$\begin{gathered}
f'(x) = \sqrt[3]{{x^2 }} - \sqrt {x^9 } = x^{\frac{2}
{3}} - x^{\frac{9}
{2}} \hfill \\
f(x) = \frac{3}
{5}x^{\frac{5}
{3}} - \frac{2}
{{11}}x^{\frac{{11}}
{2}} + C \hfill \\
\end{gathered}$
• Nov 16th 2008, 03:58 PM
endiv
Oh, wow. My mistake was putting a little c instead of a capital C, which resulted my in a wrong answer :|. Thanks!
• Nov 16th 2008, 03:59 PM
Jhevon
Quote:

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