Question

In Exercises 69 and 70 , evaluate the limit and sketch the graph of the region whose area is represented by the limit. \(\lim _{\|\Delta\| 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x,\) where \(x_{i}=i / n\) and \(\Delta x=1 / n\)

Short answer

The limit value is \(\frac{1}{6}\). The sketch of the graph is a downward-opening parabola with roots at 0 and 1. The area under the graph between these points corresponds to the limit value steps.

Step-by-step solution

Interpreting the limit

The expression \(\lim _{\|\Delta\| 0} \sum_{i=1}^{n}\left(x_{i}-x_{i}^{2}\right) \Delta x,\) represents the limit of Riemann sums that approach the definite integral of the function f(x) = x - x^2 as the partition of the interval becomes finer and finer. In this case, the interval is [0,1] because the values of \(x_i = i/n\) range between 0 and 1 for \(i = 1, ..., n\). Likewise, \(\Delta x = 1/n\) defines the width of each partition in the limit.

Evaluating the limit

The limit of the Riemann sums is equal to the definite integral of the function x-x^2 over the interval [0,1]. For the limit, we can write \(\int _0 ^1 (x-x^2) dx\). Calculating this integral requires us to find the antiderivative of the function x - x^2, apply the Fundamental theorem of Calculus, and subtract the values at the endpoints of the interval. We get \[\frac{1}{2}x^2 - \frac{1}{3}x^3 \Biggr\rvert_0 ^1 = \frac{1}{2}*1^2 - \frac{1}{3}*1^3 - (\frac{1}{2}*0^2 - \frac{1}{3}*0^3) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]

Sketching the graph

Now, we sketch the graph of f(x) = x - x^2. This is a quadratic function with roots at x = 0 and x = 1. The vertex is at x = 0.5, which is the midpoint of the roots. The graph opens downwards because the coefficient of \(x^2\) is negative. The area under the graph between x = 0 and x = 1 represents the value of the integral we have computed.