Question #4e56f

Jun 19, 2016

$\int {x}^{2} \mathrm{dx} = {x}^{3} / 3 + C$

Explanation:

Integrating any power of $x$ (such as ${x}^{2}$, ${x}^{3}$, ${x}^{4}$, and so on) is relatively straight-forward: it is done using the reverse power rule.

Recall from differential calculus that the derivative of a function like ${x}^{2}$ can be found using a handy shortcut. First, you bring the exponent to the front:
$2 {x}^{2}$
and then you decrease the exponent by one:
$2 {x}^{2 - 1} = 2 x$

Since integration is essentially the opposite of differentiation, integrating powers of $x$ should be the opposite of deriving them. To make this more clear, let's write down the steps for differentiating ${x}^{2}$:
1. Bring the exponent to the front and multiply it by $x$.
2. Decrease the exponent by one.

Now, let's think about how to do this in reverse (because integration is reverse differentiation). We need go backwards, starting at step 2. And since we're reversing the process, instead of decreasing the exponent by $1$, we need to increase the exponent by $1$. And after that, instead of multiplying by the exponent, we need to divide by the exponent. So, our steps are:
1. Increase the power by $1$.
2. Divide by the new power.

Therefore, if we need to integrate ${x}^{2}$, we increase the power by $1$:
${x}^{3}$
And divide by the new power:
${x}^{3} / 3$

All that's left is to add a constant of integration $C$ (which is done after every integration), and you're finished:
$\int {x}^{2} \mathrm{dx} = {x}^{3} / 3 + C$