amath. 數學難題

9(a) Mean value

= 1/(4 - 2) ∫ 4x^2 dx

= (1/2) (4x^3/3) |[2,4]

= (1/2) (256/3 - 32/3)

= 224/6

= 112/3

~ 37.3333

If f(x) is integrable on [a, b], then the mean value of f(x) on [a, b] is defined to be

1 b
f(x)mean = -------- ∫ f(x) dx
b – a a

f(x) = 4x^3
[2, 4]
Integrate this function
∫ 4x^3 dx = [4/(3+1)] x^(3+1) + C = x^4 + C

antiderivative = x^4 + C where C is a constant

mean value = 1/(4-2) {[(4)^4 +C] – [(2)^4 +C]}
mean value = (1/2)[256 – 16]
mean value = (1/2) 240
mean value = 120