https://file.aiquickdraw.com/mathbot/file/1725387772819mt489f4s.jpg

https://file.aiquickdraw.com/mathbot/file/1725387772819x5jilb4l.jpg

### Part (a) - Finding the Area Under the Curve

The problem asks to use **Definition 2** to find an expression for the area under the curve \( y = x^2 \) from \( 0 \) to \( 1 \) as a limit. 

#### Definition 2 (Riemann Sum):
The area under the curve \( y = f(x) \) from \( a \) to \( b \) can be approximated by the sum of the areas of rectangles:

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

where:
- \( \Delta x = \frac{b - a}{n} \) is the width of each rectangle.
- \( x_i = a + i \Delta x \) is the \(i\)-th sample point.

For the given curve \( y = x^2 \) from \( 0 \) to \( 1 \):
- \( f(x) = x^2 \)
- \( a = 0 \)
- \( b = 1 \)

Therefore, \( \Delta x = \frac{1 - 0}{n} = \frac{1}{n} \).

The sample points are \( x_i = \frac{i}{n} \), and the sum becomes:

\[
A = \lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \frac{1}{n}
\]

This simplifies to:

\[
A = \lim_{n \to \infty} \frac{1}{n^3} \sum_{i=1}^{n} i^2
\]

The formula for the sum of the squares of the first \( n \) integers is:

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

Substituting this into the expression for \( A \), we get:

\[
A = \lim_{n \to \infty} \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}
\]

\[
A = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}
\]

As \( n \) approaches infinity, the dominant terms are \( n^2 \), so:

\[
A = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}
\]

So, the area under the curve is \( \frac{1}{3} \).

### Part (b) - Using the Sum of Cubes Formula
The problem provides a formula for the sum of the cubes of the first \( n \) integers:

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

Although this formula isn't directly applicable to the sum \( \sum_{i=1}^{n} i^2 \) needed in part (a), it is related to the evaluation of similar limits. The sum of squares formula used in part (a) is distinct and necessary for the area calculation.

### Summary:
- **Part (a):** The area under the curve \( y = x^2 \) from 0 to 1 is \( \frac{1}{3} \).
- **Part (b):** Involves using the sum of cubes formula, but the sum of squares formula is what applies to this specific problem.

Would you like a more detailed explanation or have any questions?

Here are some related questions:
1. How is the Riemann sum method applied to different types of functions?
2. What is the difference between the sum of squares and the sum of cubes formulas?
3. Can you apply the Riemann sum definition to find the area under a different curve, like \( y = x^3 \)?
4. How does the limit process help in finding exact areas under curves?
5. What are some other methods to calculate areas under curves besides Riemann sums?

**Tip:** Understanding how to break down the summation formulas is key to mastering integral calculus.

Use Definition 2 to find an expression for the area under the curve y = x^2 from 0 to 1 as a limit. Then use the formula for the sum of the cubes of the first n integers to evaluate the limit.

This problem involves using Definition 2, the Riemann sum, to find the area under the curve y = x^2 from 0 to 1. The solution explores limits, sum of squares, and introduces the sum of cubes formula. Suitable for students in Calculus I.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation