Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-16, evaluate the definite integrals using the definition, as in Examples 3 and \(4 .\) \(\int_{0}^{2}\left(x^{2}+1\right) d x\) Hint: Use \(\bar{x}_{i}=2 i / n\).

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{14}{3} \).

Step 1: Understand the Definite Integral

The given definite integral is \( \int_{0}^{2} (x^2 + 1) \, dx \). This represents the area under the curve \( x^2 + 1 \) from \( x = 0 \) to \( x = 2 \).

Step 2: Set Up the Riemann Sum

To evaluate the integral using the definition, we first set up a Riemann sum. We divide the interval \([0, 2]\) into \( n \) subintervals of equal width \( \Delta x = \frac{2}{n} \). The sample point for the \( i \)-th subinterval is given as \( \bar{x}_i = \frac{2i}{n} \).

Step 3: Write the Riemann Sum Expression

The Riemann sum for the function \( x^2 + 1 \) is given by:\[S_n = \sum_{i=1}^{n} \left(\left(\frac{2i}{n}\right)^2 + 1\right) \cdot \frac{2}{n}\]This expression will be simplified and evaluated to find the limit as \( n \to \infty \).

Step 4: Substitute and Simplify the Function Values

Substitute \( \bar{x}_i = \frac{2i}{n} \) into the function \( x^2 + 1 \):\[\left(\frac{2i}{n}\right)^2 + 1 = \frac{4i^2}{n^2} + 1\]Substitute this into the Riemann sum expression:\[S_n = \sum_{i=1}^{n} \left(\frac{4i^2}{n^2} + 1\right) \cdot \frac{2}{n}\]\[= \sum_{i=1}^{n} \left(\frac{8i^2}{n^3} + \frac{2}{n}\right)\]

Step 5: Separate the Riemann Sum Expressions

Split the Riemann sum into two separate sums:\[S_n = \sum_{i=1}^{n} \frac{8i^2}{n^3} + \sum_{i=1}^{n} \frac{2}{n}\]

Step 6: Evaluate Each Sum Using Known Formulas

Use the formula for the sum of squares: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \):\[\sum_{i=1}^{n} \frac{8i^2}{n^3} = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}\]For the second sum, we have:\[\sum_{i=1}^{n} \frac{2}{n} = \frac{2n}{n} = 2\]

Step 7: Take the Limit as \( n \to \infty \)

Evaluate the limits of the two sums as \( n \) approaches infinity:\[\lim_{n \to \infty} \left( \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + 2 \right)\]The first term simplifies to \( \frac{8}{6} \cdot 2 = \frac{8}{3} \), and the second term is \( 2 \). Thus, the entire limit evaluates to:\[\frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}\]

Step 8: Conclusion

The value of the definite integral \( \int_{0}^{2} (x^2 + 1) \, dx \) is \( \frac{14}{3} \).

Key Concepts

Riemann sum

A Riemann sum is a method for estimating the total area under a curve on a graph, otherwise known as the integral of a function. To approximate this area using a Riemann sum, the interval \([a, b]\) is divided into smaller subintervals. The width of these subintervals is denoted by \(\Delta x\), and a sample point \(\bar{x}_i\) is chosen within each subinterval. For our problem, the interval \([0, 2]\) is divided into \(n\) subintervals of equal width \(\Delta x = \frac{2}{n}\). The sample point in the \(i\)-th subinterval is \(\bar{x}_i = \frac{2i}{n}\). By substituting this sample point into the function \((x^2 + 1)\), we form the Riemann sum. This initial step creates the following expression:

• \(S_n = \sum_{i=1}^{n} \left(\left(\frac{2i}{n}\right)^2 + 1\right) \cdot \frac{2}{n}\)

By evaluating this sum as \(n\) approaches infinity, we can find the exact area under the curve.

limit evaluation

Limit evaluation in the context of definite integrals involves finding the value of a sum as the number of subintervals, \(n\), approaches infinity. This process essentially narrows down the calculation error in approximating the area under the curve using the Riemann sum.When calculating this limit, we first simplify the Riemann sum. After determining the separate sums, we apply limit principles:

• For \(+\lim_{n \to \infty} \sum_{i=1}^{n} \frac{8i^2}{n^3}\), recognize that we will substitute with the sum of squares formula and proceed to the limit.
• The constant term sum, \(+\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n}\), leads directly to a constant value of 2.

Achieving these limits produces the precise result of the definite integral.

sum of squares formula

The sum of squares formula is a crucial mathematical tool when evaluating limits of sums. It assists in breaking down and simplifying expressions involving squares of integers. For sequence sums like \( \sum_{i=1}^{n} i^2 \), the formula is:\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]By substituting this formula into the Riemann sum expression, we simplify it:

• Consider \(+\frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}\) from our problem.

This leads to a simpler expression that is easier to handle when \(n\) approaches infinity. The formula provides an exact technique to compute these sums systematically, easing the transition into taking limits.

area under the curve

Calculating the area under the curve of a function is an essential component of definite integrals. This area represents the integral of the function across a certain interval. In this case, the integral \(\int_{0}^{2} (x^2 + 1) \, dx\) represents the area under the curve \((x^2 + 1)\) from \(x=0\) to \(x=2\).The entire process from setting up the Riemann sum, simplifying with the sum of squares, and evaluating the limits is focused on accurately calculating this area. In our problem, the final expression of the integral \(= \frac{14}{3}\) is the exact measurement of the area under the curve, demonstrating the effectiveness of definite integrals in solving such geometrical problems.