Vtyjkyuk

I have to calculate $$int_-1^1|x^n| dx$$

I don't know how to delete the absolute value.

Do I write $-x^n$ where $x$ from $-1$ to $0$ or write $(-1)^n x^n$ ?

Thank you

The integrand is even, so the integral is $2int_0^1 x^n dx=frac2n+1$ for $n>-1$.

You have $|(-x)^n|=|x^n|$ so $int_-1^0 |x^n|dx = int_0^1 |x^n|dx = int_0^1 x^n dx$.

Here $|x^n|=x^n$ for $xge 0$ because $x^nge 0$.

You just have to consider that $;left | x^n right | = left | x right |^n$

Then, undo the absolute value as usual:
$$left | x right |=left{beginmatrixx,& &xgeq0&\ -x,& & x<0&endmatrixright.$$

Use the additivity of integration on intervals to compute:

$$int_-1^1left | x^n right |dx=int_-1^0(-1)^nx^ndx;+;int_0^1x^ndx,quad forall n>0qquad left(1right)$$

Be careful!

For $n<0,$ the function is not bounded. It is a simple example of improper integral.

However, for $,-1<n<0,$ the improper integral converges so you can be sure that $,left(1right),$ is true $,forall n > -1.$

The particular case $,n=0,$ has no mathematical interest cause $,left|xright|^n=1,$ if $,xneq0,$ and $,displaystylelim_xto 0,left|xright|^x = 1.$