karush
Gold Member
MHB
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$\tiny{s8.5.5.2}$
Find the average value of the function on the given interval. $\sqrt{x}\quad [0,4]$
average value $\boxed{f_{ave}=\dfrac{1}{b-a}\int_a^b f(x) \ dx}$
so with $a=0$ and $b=4$ thus
$\dfrac{1}{4-0}\displaystyle\int_0^4 \sqrt{x} \ dx
\implies =\dfrac{1}{4}\left[\dfrac{2}{3}x^{\dfrac{3}{2}}\right]^4_0
\implies = \dfrac{1}{4}\dfrac{16}{3}=\dfrac{4}{3}$
ok, I think this is correct, but possible typos
I want to poIst this problem on Linkedin this week so if there is any added tips
i will add it in... been getting lots of views on IN even tho it is not essentually a math forum
of I Suggest they come here
Find the average value of the function on the given interval. $\sqrt{x}\quad [0,4]$
average value $\boxed{f_{ave}=\dfrac{1}{b-a}\int_a^b f(x) \ dx}$
so with $a=0$ and $b=4$ thus
$\dfrac{1}{4-0}\displaystyle\int_0^4 \sqrt{x} \ dx
\implies =\dfrac{1}{4}\left[\dfrac{2}{3}x^{\dfrac{3}{2}}\right]^4_0
\implies = \dfrac{1}{4}\dfrac{16}{3}=\dfrac{4}{3}$
ok, I think this is correct, but possible typos
I want to poIst this problem on Linkedin this week so if there is any added tips
i will add it in... been getting lots of views on IN even tho it is not essentually a math forum
of I Suggest they come here