# Thread: Calculate the average over the given interval.

1. ## Calculate the average over the given interval.

I do not understand the final step taken to solve this problem (It was an example given in class) I would really appreciate it if someone would be able to explain the simplification process for the final step?

f
(x) = xn
for
n ≥ 0, [0, 2]

${\frac{1}{2-0} }\int\limits_{0}^{2}x^{n}{dx}$ = ${\frac{1}{2n + 2}x^{n+1} \Bigr\rvert\biggr \limits_{0}^{2}$

I understand that one must solve using the given limits, which would give me a value of:

$\biggl[{\frac{1}{2n + 2}2^{n+1} \biggr]$ - $\biggl[{\frac{1}{2n + 2}0^{n+1} \biggr]$

However, I am not following the what happens in order to have the final solution of:

${2^{n}\frac{1}{n+1}}$

I end up with my final solution being:
${\frac{1}{2n + 2}2^{n+1}$

Any sort of help is greatly appreciated! Thank you

2. ## Re: Calculate the average over the given interval.

Originally Posted by cathrinbleu
I do not understand the final step taken to solve this problem (It was an example given in class) I would really appreciate it if someone would be able to explain the simplification process for the final step?

f(x) = xn
for
n ≥ 0, [0, 2]

${\frac{1}{2-0} }\int\limits_{0}^{2}x^{n}{dx}$ = ${\frac{1}{2n + 2}x^{n+1} \Bigr\rvert\biggr \limits_{0}^{2}$

I understand that one must solve using the given limits, which would give me a value of:

$\biggl[{\frac{1}{2n + 2}2^{n+1} \biggr]$ - $\biggl[{\frac{1}{2n + 2}0^{n+1} \biggr]$

However, I am not following the what happens in order to have the final solution of:

${2^{n}\frac{1}{n+1}}$

I end up with my final solution being: ${\frac{1}{2n + 2}2^{n+1}$
${\frac{1}{2n + 2}2^{n+1}={\frac{1}{2(n + 1)}2(2^{n})=~?$

3. ## Re: Calculate the average over the given interval.

I suppose I am not following where you take the 2 out of $2^{n+1}$ in order to get $2(2^{n})$

I imagine it's quite simple, yet for some reason it's lost on me....how frustrating

4. ## Re: Calculate the average over the given interval.

Originally Posted by cathrinbleu
I suppose I am not following where you take the 2 out of $2^{n+1}$ in order to get $2(2^{n})$
$2(2^n)=2^1(2^n)=2^{n+1}$