Calculus I
Why is the antidervative of
x^ n = [x^(n+1)]/ (n+1) + C]?

i.e if I have x^2 the derivative is 2x
why is the integral of
2x = [(x^2)/2] + C ?

If I differentiate the (x^2)/2 I have to use the quotient rule and I don't get 2x which is what I started with.

Quotient Rule (f'g-fg')/g^2

Question

Calculus I
Why is the antidervative of 
x^ n = [x^(n+1)]/ (n+1) + C]?

i.e if I have x^2 the derivative is 2x
why is the integral of 
2x = [(x^2)/2] + C ?

If I differentiate the (x^2)/2 I have to use the quotient rule and I don't get 2x which is what I started with.

Quotient Rule                 (f'g-fg')/g^2

anonymous · Answer

I read the definition but it doesn't make sense to me.
The antidervative is supposed to undo the derivative.

anonymous · Answer

$$
\int x\ dx=\frac{1}{2}x^2+C\
\int 2x\ dx=2\int x\ dx = 2\cdot\frac{1}{2}x^2+C=x^2+C

$$

anonymous · Answer

$$
\int ax^n\ dx=\frac{a}{n+1}x^{n+1}+C
$$
You're just dropping a 2 somewhere and that's why your results are getting mixed up.

anonymous · Answer

the dervative of $$x^2 = 2x$$
so why do we divided by 2

anonymous · Answer

or multiply by 1/2

anonymous · Answer

nvm I found where I messed up. Thanks for all the help.