If the function f given by f(x)=x3 has an average value of 9 on the closed interval [0,k],then k=?

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If the function f given by f(x)=x3 has an average value of 9 on the closed interval [0,k],then k=?

If the function f given by f(x)=x3 has an average value of 9 on the closed interval [0,k],then k=?
If the function f given by f(x)=x3 has an average value of 9 on the closed interval [0,k],then k=?

If the function f given by f(x)=x3 has an average value of 9 on the closed interval [0,k],then k=?
f(x)=x^3?
[∫f(x)dx]/k=9
k=36^(1/3)

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