1.\nif the function ( f ) given by ( f(x)=x^{3} ) has an average value of 9 on the closed interval (0, k)…

1.\nif the function ( f ) given by ( f(x)=x^{3} ) has an average value of 9 on the closed interval (0, k), then ( k= )\n(a) 3\n(b) ( 3^{\frac{1}{2}} )\n(c) ( 18^{\frac{1}{3}} )\n(d) ( 36^{\frac{1}{4}} )

1.\nif the function ( f ) given by ( f(x)=x^{3} ) has an average value of 9 on the closed interval (0, k), then ( k= )\n(a) 3\n(b) ( 3^{\frac{1}{2}} )\n(c) ( 18^{\frac{1}{3}} )\n(d) ( 36^{\frac{1}{4}} )

Answer

Explanation:

Step1: Recall the formula for the average value of a function

The average value of a function (y = f(x)) on the interval ([a,b]) is given by (f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx). Here, (a = 0), (b=k), and (f(x)=x^{3}), so (f_{avg}=\frac{1}{k-0}\int_{0}^{k}x^{3}dx).

Step2: Compute the integral

We know that (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). So, (\int_{0}^{k}x^{3}dx=\left[\frac{x^{4}}{4}\right]{0}^{k}=\frac{k^{4}}{4}-0=\frac{k^{4}}{4}). Then (f{avg}=\frac{1}{k}\cdot\frac{k^{4}}{4}=\frac{k^{3}}{4}).

Step3: Set up the equation and solve for (k)

Since (f_{avg}=9), we have the equation (\frac{k^{3}}{4}=9). Multiply both sides by (4) to get (k^{3}=36). Then take the cube - root of both sides: (k = 36^{\frac{1}{3}}).

Answer:

D. (36^{\frac{1}{4}})