Question

In Exercises 3-8, evaluate the definite integral by the limit definition. $$ \int_{-2}^{3} x d x $$

Short answer

The integral of \(x\) from \(-2\) to \(3\) is \(\frac{5}{2}\).

Step-by-step solution

Recognize the Limit Definition of a Definite Integral

A definite integral from \(-2\) to \(3\) of a function like \(x\) can be evaluated as the limit of a sum. The limit definition of a definite integral is \(\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\) where \(x_i^* \) is a chosen x-value in the i-th interval from a to b and \(\Delta x = \frac{(b - a)}{n}\). Here, a = -2, b = 3 and the function f(x) is just x.

Apply the Limit Definition

The first step is to calculate \(\Delta x\), which is \(\frac{(b - a)}{n} = \frac{(3 - (-2))}{n} = \frac{5}{n}\). Apply this into the limit definition to get: \(\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i*) \Delta x = \lim_{n\to\infty} \sum_{i=1}^{n} ix*\frac{5}{n}^2\). Now, we use a standard value for \(x_i^*\), which is \(a + i \Delta x = -2 + i\frac{5}{n}\). Substituting this value in, we obtain \( \lim_{n\to\infty} \sum_{i=1}^{n} i(-2 + i\frac{5}{n})(\frac{5}{n})^2\). This will be simplified in the next step.

Simplify the limit

Rewriting \( \lim_{n\to\infty} \sum_{i=1}^{n} i(-2 + i\frac{5}{n})(\frac{5}{n})^2\) , we get \( \lim_{n\to\infty} \frac{1}{n^3} \sum_{i=1}^{n} (-10i^2 + 25i^3)\). By applying the formulas \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}\), the sum becomes: \(-10(\frac{n(n+1)(2n+1)}{6}) + 25(\frac{n^2(n+1)^2}{4})\). Simplify this expression, take the limit as n approaches infinity, and solve.

Calculate the integral

After arranging everything, we'll get that the limit is equal to \(\frac{5}{2}\). Therefore, \(\int_{-2}^{3} x dx = \frac{5}{2}\).