f ave of a function

Question

Find the exact average value  of  between 0 and 3, then find an exact number  in [0,3] where .

f ave=

c=

Answer 1

Average value = 1/(3-0)∫_{(from 0 to 3)} √(9 - x²)dx

The graph of y = √(9 - x²) where x is in [0,3] is the upper right quarter circle of the circle with center (0,0) and radius 3. So, the integral is merely the area of this quarter circle.

Thus, Average value = (1/3)(9π/4) = 3π/4

So, find c in [0,3] so that √(9-x²) = 3π/4:

9 - x² = 9π²/16

x² = 9 - 9π²/16

x = 3√ (1-π²/16) = (3/4)√(16 - π²)

Answer 2

f_ave = (f(3)+f(0))/(3-0) = (0+√9)/3 = 1

√(9-c²) = 1

9-c²=1

c = √(9-1) = 2√2