Easy Integral

**fatmoldy** · December 1st 2009, 08:00 PM

I can't remember how to do these stupid integrals... The one im stuck on is
x-x^2. It seems like it should be easy but im not getting anywhere.

**RockHard** · December 1st 2009, 08:08 PM

$\int x-x^2 = \frac{x^2}{2} - \frac{x^3}{3}$

You add 1 to the exponents and then divide that term by the new raised power

**fatmoldy** · December 1st 2009, 08:14 PM

Originally Posted by RockHard

$\int x-x^2 = \frac{x^2}{2} - \frac{x^3}{3}$

You add 1 to the exponents and then divide that term by the new raised power

I forgot you could split them up. Thanks.

**Bruno J.** · December 1st 2009, 08:31 PM

Originally Posted by RockHard

$\int x-x^2 = \frac{x^2}{2} - \frac{x^3}{3}$

You add 1 to the exponents and then divide that term by the new raised power

Good, but be careful with your notation! It should be $\int (x-x^2)dx$ , not just $\int x-x^2$ which has no meaning. The $dx$ is there for a reason! It might seem unnecessary when you begin calculus but later on you'll have a lot of trouble keeping up if you're not careful with notation.

**Drexel28** · December 1st 2009, 08:47 PM

Originally Posted by Bruno J.

Good, but be careful with your notation! It should be $\int (x-x^2)dx$ , not just $\int x-x^2$ which has no meaning. The $dx$ is there for a reason! It might seem unnecessary when you begin calculus but later on you'll have a lot of trouble keeping up if you're not careful with notation.

Who knows, maybe he could have been wanting to compute $\int_0^1\left(x-x^2\right)d\sin(x)$ :O

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