need help solving this calc question

Question

 

Find the average value of f(x) in the interval [-1,3] when f'(x)=3x^2-6x and f(2)=0?

Answer 1

The avarage value of f(x) in the interval [-1,3]

When

f(2) = ?

Note:   Here

Given that interval [-1,3]

There fore

Formula:

Formula:

Find f(2) = ?

Formula:

Comments

The average value of f (x ) in the interval [-1, 3], f_average  = 2.

Answer 2

f '(x ) = 3x² - 6x

Take integration to each side.

ʃ f '(x ) = ʃ (3x² - 6x ) dx

f (x ) = ʃ (3x² - 6x ) dx

= ʃ (3x²) dx - ʃ (6x ) dx

= 3ʃ x² dx - 6ʃx dx

= 3(x³/3)  - 6(x²/2) + C

= x³  - 3x² + C -> (1)

Condition is f(2) = 0.

0 = (2)³  - 3(2)² + C

0 = 8  - 3(4) + C

0 = 8  - 12 + C

0 = -4 + C

C = 4

Substitute c value in(1)

f (x ) = x³  - 3x² + 4

The average value of the function f (x ) over the interval [a, b] is .

Therefore the average value of f (x ) in the interval [-1, 3], f_average  = 2.