Question

Evaluate \(\int\limits_{2}^3x^2\ dx\) as the limit of a sum

• A. \(\frac{72}{6}\)
• B. \(\frac{25}{7}\)
• C. \(\frac{53}{9}\)
• D. \(\frac{19}{3}\)

Answer 1

The Correct Option is D

We are asked to evaluate the definite integral \[ \int\limits_{2}^3 x^2 \, dx \] as the limit of a sum. 

Step 1: Partition the interval
Let the interval \([2, 3]\) be divided into \(n\) subintervals, each of width \[ \Delta x = \frac{3 - 2}{n} = \frac{1}{n} \] 

Step 2: Choose sample points
Using right endpoints: \[ x_i = 2 + i \cdot \frac{1}{n}, \quad \text{for } i = 1, 2, \dots, n \] 

Step 3: Define the Riemann sum
\[ \sum_{i=1}^{n} f(x_i) \cdot \Delta x = \sum_{i=1}^{n} \left(2 + \frac{i}{n}\right)^2 \cdot \frac{1}{n} \] 

Step 4: Take the limit
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \left(2 + \frac{i}{n}\right)^2 \cdot \frac{1}{n} = \int_{2}^{3} x^2 \, dx \] 

Step 5: Compute the integral
\[ \int_{2}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_2^3 = \frac{3^3}{3} - \frac{2^3}{3} = \frac{27 - 8}{3} = \frac{19}{3} \] 

Final Answer: \(\frac{19}{3}\)

Answer 2

We want to evaluate the definite integral \(\int\limits_{2}^3 x^2 \, dx\) using the definition of the integral as the limit of a Riemann sum.

The definition of the definite integral is: 

\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x \]

In this problem, we have:

• The function \(f(x) = x^2\)
• The interval \([a, b] = [2, 3]\), so \(a = 2\) and \(b = 3\)

First, calculate the width of each subinterval:

\[ \Delta x = \frac{b - a}{n} = \frac{3 - 2}{n} = \mathbf{\frac{1}{n}} \]

Next, determine the sample points \(x_i\). We will use the right endpoints of the subintervals:

\[ x_i = a + i \Delta x = 2 + i \left(\frac{1}{n}\right) = \mathbf{2 + \frac{i}{n}} \]

Now, evaluate the function \(f(x)\) at these sample points:

\[ f(x_i) = f\left(2 + \frac{i}{n}\right) = \left(2 + \frac{i}{n}\right)^2 \]

\[ f(x_i) = 2^2 + 2(2)\left(\frac{i}{n}\right) + \left(\frac{i}{n}\right)^2 = \mathbf{4 + \frac{4i}{n} + \frac{i^2}{n^2}} \]

Now, form the Riemann sum:

\[ \sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n \left(4 + \frac{4i}{n} + \frac{i^2}{n^2}\right) \left(\frac{1}{n}\right) \]

\[ = \frac{1}{n} \sum_{i=1}^n \left(4 + \frac{4i}{n} + \frac{i^2}{n^2}\right) \]

Distribute the summation:

\[ = \frac{1}{n} \left[ \sum_{i=1}^n 4 + \sum_{i=1}^n \frac{4i}{n} + \sum_{i=1}^n \frac{i^2}{n^2} \right] \]

Factor out constants from the summations:

\[ = \frac{1}{n} \left[ 4 \sum_{i=1}^n 1 + \frac{4}{n} \sum_{i=1}^n i + \frac{1}{n^2} \sum_{i=1}^n i^2 \right] \]

Use the standard summation formulas:

• \(\sum_{i=1}^n 1 = n\)
• \(\sum_{i=1}^n i = \frac{n(n+1)}{2}\)
• \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\)

Substitute these formulas into the expression:

\[ = \frac{1}{n} \left[ 4(n) + \frac{4}{n} \left(\frac{n(n+1)}{2}\right) + \frac{1}{n^2} \left(\frac{n(n+1)(2n+1)}{6}\right) \right] \]

Simplify the terms:

\[ = \frac{1}{n} \left[ 4n + 2(n+1) + \frac{(n+1)(2n+1)}{6n} \right] \]

Distribute the \(\frac{1}{n}\) outside the bracket:

\[ = \frac{4n}{n} + \frac{2(n+1)}{n} + \frac{(n+1)(2n+1)}{6n^2} \]

\[ = 4 + 2\left(\frac{n+1}{n}\right) + \frac{2n^2 + 3n + 1}{6n^2} \]

\[ = 4 + 2\left(1 + \frac{1}{n}\right) + \left(\frac{2n^2}{6n^2} + \frac{3n}{6n^2} + \frac{1}{6n^2}\right) \]

\[ = 4 + 2\left(1 + \frac{1}{n}\right) + \left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right) \]

Finally, take the limit as \(n \to \infty\):

\[ \int_2^3 x^2 \, dx = \lim_{n \to \infty} \left[ 4 + 2\left(1 + \frac{1}{n}\right) + \left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right) \right] \]

As \(n \to \infty\), the terms \(\frac{1}{n}\), \(\frac{1}{2n}\), and \(\frac{1}{6n^2}\) all approach 0.

\[ = 4 + 2(1 + 0) + \left(\frac{1}{3} + 0 + 0\right) \]

\[ = 4 + 2(1) + \frac{1}{3} \]

\[ = 4 + 2 + \frac{1}{3} \]

\[ = 6 + \frac{1}{3} \]

\[ = \frac{18}{3} + \frac{1}{3} = \mathbf{\frac{19}{3}} \]

Comparing this with the given options, the correct option is:

\(\frac{19}{3}\)