Let's say I have function $f(x) = 2^{x-2}+1$. I now have to detirmine the value of $a$, such that the average value of the function $f(x)$ on the interval $[2,a]$ is 8. This number doesn't have the be exact, so a decimal would be ok.
I already know that the average value of the function over the interval $[a,b]$ would be $(1/{b-a})*\int^b_af(x)dx$. So that means I have to solve this integral but I'm not sure how: $1/(a-2)\int^a_2(2^{x-2}+1)dx$.
in $\frac{1}{a-2}\int^a_2(2^{x-2}+1)dx $ , the primitive is $x+\frac{2^{x-2}}{log(2)}$
Then we know that :
$\frac{1}{a-2} ( ( a+\frac{2^{a-2}}{log(2)} ) - ( 2+\frac{1}{log(2)} ) ) = 8$
then
$\frac{1}{a-2} ( ( a+\frac{2^{a-2}}{log(2)} ) - ( 2+\frac{2^{2-2}}{log(2)} ) ) = 8$ => $a \approx 6.51$