Question

(a) Find the average value of \({\rm{f}}\) on the given interval.

(b) Find \({\rm{c}}\) such that \({{\rm{f}}_{{\rm{ave}}}}{\rm{ = f(c)}}\)

(c) Sketch the graph of \({\rm{f}}\)and a rectangle whose area is the same as the area under the graph of \({\rm{f}}\)

\({\rm{f(x) = }}\sqrt {\rm{x}} {\rm{,(0,4)}}\)\({\rm{f(x) = }}\sqrt {\rm{x}} {\rm{,(0,4)}}\)

Short answer

(a)\(\frac{{\rm{4}}}{{\rm{3}}}\) (b) \({\rm{c = }}\frac{{{\rm{16}}}}{{\rm{9}}}\) (c) the rectangle with base \({\rm{(0,4)}}\) has the same area

Step-by-step solution

Step 1: the average value of\({\rm{f}}\).Step 1: the average value of\({\rm{f}}\).

In general, the average value of the function\({\rm{f}}\)on the interval \({\rm{(a,b)}}\)is

(a) Consider the function\({\rm{f(x) = }}\sqrt {\rm{x}} \)

And the interval \({\rm{(0,4)}}\)

Now, find the average value of this function. Here\({\rm{a = 0,b = 4}}\)

So,

\(\begin{aligned}{c}{\rm{ &= }}\frac{{\rm{1}}}{{\rm{6}}}\left( {{{\rm{4}}^{{\rm{3/2}}}}{\rm{ - 0}}} \right)\\{\rm{ &= }}\frac{{\rm{1}}}{{\rm{6}}}{\rm{ \times 8 &= }}\frac{{\rm{4}}}{{\rm{3}}}\end{aligned}\)

Therefore, the average value of this function is \(\frac{{\rm{4}}}{{\rm{3}}}\)

Finding \({\rm{c}}\)

(b) Equating\({\rm{f(c)}}\)to the average value of\({\rm{f}}\),

Here,

\(\begin{aligned}{c}{\rm{f(c) &= }}{{\rm{f}}_{{\rm{ave\;}}}}\\\frac{{\rm{4}}}{{\rm{3}}}{\rm{ &= }}\sqrt {\rm{c}} \\{\rm{c &= }}\frac{{{\rm{16}}}}{{\rm{9}}}\\{\rm{Thus,c &= }}\frac{{{\rm{16}}}}{{\rm{9}}}\end{aligned}\)

Sketching the graph.

(c) Now, sketch the graph of \({\rm{f(x) = }}\sqrt {\rm{x}} \)and a rectangle whose area is the same as the area under the graph of\({\rm{f}}\).

Recall that,

From the Mean Value Theorem of integrals, if\({\rm{f}}\)is a positive function and there is a number\({\rm{c}}\)such that the rectangle with base\({\rm{(a,b)}}\)and height\({\rm{f(c)}}\)has the same area as the region under the graph of\({\rm{f}}\)from\({\rm{a to b}}\).

From this\({\rm{f(x) = }}\sqrt {\rm{x}} \)is a positive function..

Therefore, the rectangle with base \({\rm{(0,4)}}\) has the same area of the graph of \({\rm{f(x) = }}\sqrt {\rm{x}} \)