Answer
$1.9375$
Work Step by Step
Given that $ƒ(x) = x^3$ on the interval $[0, 2]$.
Now, using a finite sum, we can partition the interval into four subintervals of equal length and evaluate ƒ at the subinterval midpoints.
The length of each subinterval can be calculated as follows:
Length of subinterval = $\dfrac{\text{upper bound} - \text{lower bound}}{\text{number of subintervals}}$
= $\frac{(2 - 0)}{4}$
= $0.5$
Now, we can calculate the subinterval midpoints:
Subinterval $1$: midpoint = $\frac{(0 + 0.5) }{2}= 0.25$
Subinterval $2$: midpoint = $\frac{(0.5 + 1)}{2}= 0.75$
Subinterval $3$: midpoint = $\frac{(1 + 1.5)}{2}= 1.25$
Subinterval $4$: midpoint = $\frac{(1.5 + 2)}{2}= 1.75$
Next, we evaluate ƒ at these subinterval midpoints:
$ƒ(0.25) = (0.25)^{3}= 0.015625$
$ƒ(0.75) = (0.75)^{3}= 0.421875$
$ƒ(1.25) = (1.25)^{3} = 1.953125$
$ƒ(1.75) = (1.75)^{3} = 5.359375$
Finally, we can calculate the average value of ƒ on the interval $[0, 2]$ using the finite sum:
Average value of ƒ = $\frac{\text{sum of function values}}{\text{number of subintervals}}$
= $\frac{(0.015625 + 0.421875 + 1.953125 + 5.359375)}{4}$
= $\frac{ 7.75 }{4}$
= $1.9375$
