(a) Use the definition given below with right endpoints to express the area under the curve y = x3 from 0 to 1 as a limit. The area A of the region S that is bounded above by the graph of a continuous function y = f(x), below by the x-axis, and on the sides by the lines x = a and x = b is the limit of the sum of the areas of approximating rectangles. A = lim n→∞ Rn = lim n→∞ [f(x1)Δx + f(x2)Δx + + f(xn)Δx] = lim n→∞ n i=1 f(xi)Δx lim n→∞ n i = 1
(b) Use the following formula for the sum of cubes of the first n integers to evaluate the limit in part (a). 13 + 23 + 33 + + n3 = n(n + 1) 2 2

We are asked to express the area under the curve \( y = x^3 \) from 0 to 1 using the right endpoint approximation and then evaluate this limit using the sum of cubes formula.

### Part (a): Express the area as a limit

The general formula for the area under a curve using a Riemann sum with right endpoints is given by:

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

Where:
- \( \Delta x \) is the width of each rectangle.
- \( x_i \) is the right endpoint of each subinterval.
- \( f(x_i) \) is the value of the function at the right endpoint.

For the function \( f(x) = x^3 \) over the interval from 0 to 1:
- \( \Delta x = \frac{1-0}{n} = \frac{1}{n} \)
- The right endpoint of the \(i\)-th subinterval is \( x_i = \frac{i}{n} \), where \( i \) runs from 1 to \( n \).

Thus, the Riemann sum for the area under \( y = x^3 \) is:

\[
A_n = \sum_{i=1}^{n} f\left(\frac{i}{n}\right) \Delta x = \sum_{i=1}^{n} \left(\frac{i}{n}\right)^3 \cdot \frac{1}{n}
\]

Simplifying this expression:

\[
A_n = \frac{1}{n^4} \sum_{i=1}^{n} i^3
\]

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

\[
A = \lim_{n \to \infty} \frac{1}{n^4} \sum_{i=1}^{n} i^3
\]

### Part (b): Use the sum of cubes formula

The sum of cubes of the first \( n \) integers is given by:

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

Substituting this into our expression for \( A \):

\[
A = \lim_{n \to \infty} \frac{1}{n^4} \cdot \left( \frac{n(n+1)}{2} \right)^2
\]

Now, let's expand and simplify the expression inside the limit:

\[
A = \lim_{n \to \infty} \frac{1}{n^4} \cdot \frac{n^2 (n+1)^2}{4}
\]
\[
A = \lim_{n \to \infty} \frac{n^2 (n+1)^2}{4n^4}
\]

Next, simplify the fraction:

\[
A = \lim_{n \to \infty} \frac{(n+1)^2}{4n^2}
\]

Expanding \( (n+1)^2 \):

\[
A = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{4n^2}
\]

Now, divide each term by \( n^2 \):

\[
A = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4}
\]

As \( n \to \infty \), the terms \( \frac{2}{n} \) and \( \frac{1}{n^2} \) approach 0, so we are left with:

\[
A = \frac{1}{4}
\]

### Final Answer:
The area under the curve \( y = x^3 \) from 0 to 1 is \( \frac{1}{4} \).

---

Would you like further details on the steps, or do you have any questions?

Here are 5 related questions:
1. How does the concept of Riemann sums generalize to other curves?
2. Can the midpoint rule or trapezoidal rule be used for better approximations of this area?
3. How would the sum of squares formula be used for functions like \( y = x^2 \)?
4. What happens to the limit if the interval changes from \( [0, 1] \) to \( [0, 2] \)?
5. How does the limit definition relate to the Fundamental Theorem of Calculus?

**Tip**: When evaluating limits, always simplify terms carefully to ensure you catch terms that vanish as \( n \to \infty \).

Use the definition given below with right endpoints to express the area under the curve y = x^3 from 0 to 1 as a limit. The area A of the region S that is bounded above by the graph of a continuous function y = f(x), below by the x-axis, and on the sides by the lines x = a and x = b is the limit of the sum of the areas of approximating rectangles. A = lim n→∞ Rn = lim n→∞ [f(x1)Δx + f(x2)Δx + + f(xn)Δx] = lim n→∞ n i=1 f(xi)Δx lim n→∞ n i = 1. Use the following formula for the sum of cubes of the first n integers to evaluate the limit in part (a). 13 + 23 + 33 + + n3 = n(n + 1)^2/4

Learn how to express the area under the curve y = x^3 from 0 to 1 as a limit using Riemann sums. This detailed solution includes using the sum of cubes formula to evaluate the limit, showcasing key calculus concepts. Suitable for Grades 11-12 and college-level students.

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