To determine

To find:

The number b such that average value of f(x) on the interval 0,b is 3.

Answer

b=3+52=2.618  and  b=3-52=0.382

Explanation

1) Concept:

Use the mean value theorem for integrals.

2) Mean Value Theorem for Integrals

If f is continuous on a,b then there exists a number c in a,b such that

fc=fave=1b-a∫abfx dx

that is,

∫abfx dx=fcb-a

3) Given:

i) fx=2+6x-3x2

ii) fave=3

4) Calculation:

Since f is a polynomial function,

f is continuous over R.

Therefore, for any b∈R

f is continuouson the interval 0,b.

By using themean value theorem for integrals

there exists a number c in 0,b such that

fc=fave=1b-0∫0bfx dx

Using the given information,

3=1b∫0b2+6x-3x2 dx

3b=∫ob2+6x-3x2 dx

By integrating,

3b=2x+6x22-3x330b

3b=2b+6b22-3b33-2(0)+6022-3033

Simplifying,

3b=2b+6b22-3b33-0

3b-2b-6b22+3b33=0

b-3b2+b3=0

b1-3b+b2=0

Since b≠0,

b2-3b+1=0

b=--3±9-42

b=3±52

b=3+52=2.618>0  and  b=3-52=0.382>0,

so both the values are the required solution.

Therefore,

b=3+52=2.618  or  b=3-52=0.382

Conclusion:

The number

b=3+52=2.618  and  b=3-52=0.382

are such that the average value of f(x) on the interval 0,b is 3.