The problem asks us to find the overestimate of the area under the curve $f(x) = x^3$ on the interval $[0, 1]$ when the interval is divided into 3 equal parts. This means we will be finding the area of the rectangles formed by using the right endpoint of each subinterval as the height of the rectangle.

1. Problem Description

The problem asks us to find the overestimate of the area under the curve $f(x) = x^3$ on the interval $[0, 1]$ when the interval is divided into 3 equal parts. This means we will be finding the area of the rectangles formed by using the right endpoint of each subinterval as the height of the rectangle.

2. Solution Steps

The interval $[0, 1]$ is divided into 3 equal parts, so the width of each subinterval is $\Delta x = \frac{1 - 0}{3} = \frac{1}{3}$.

The subintervals are $[0, \frac{1}{3}]$, $[\frac{1}{3}, \frac{2}{3}]$, and $[\frac{2}{3}, 1]$.

Since we are looking for the overestimate, we use the right endpoint of each subinterval to determine the height of each rectangle.

The right endpoints are $\frac{1}{3}$, $\frac{2}{3}$, and $1$.

The heights of the rectangles are $f(\frac{1}{3}) = (\frac{1}{3})^3 = \frac{1}{27}$, $f(\frac{2}{3}) = (\frac{2}{3})^3 = \frac{8}{27}$, and $f(1) = 1^3 = 1$.

The area of the first rectangle is $A_1 = f(\frac{1}{3}) \cdot \Delta x = \frac{1}{27} \cdot \frac{1}{3} = \frac{1}{81}$.

The area of the second rectangle is $A_2 = f(\frac{2}{3}) \cdot \Delta x = \frac{8}{27} \cdot \frac{1}{3} = \frac{8}{81}$.

The area of the third rectangle is $A_3 = f(1) \cdot \Delta x = 1 \cdot \frac{1}{3} = \frac{1}{3} = \frac{27}{81}$.

The total area of the rectangles (the overestimate) is the sum of the individual areas:

$A = A_1 + A_2 + A_3 = \frac{1}{81} + \frac{8}{81} + \frac{27}{81} = \frac{1 + 8 + 27}{81} = \frac{36}{81} = \frac{4}{9}$.

3. Final Answer

The overestimate of the area is $\frac{4}{9}$.