Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the region under the curve \(y=f(x)\) over the interval \([a, b]\). To do this, divide the interval \([a, b]\) into n equal subintervals, calculate the area of the corresponding circumscribed polygon, and then let \(n \rightarrow \infty\). $$ y=x^{3} ; a=0, b=1 $$

Step-by-Step Solution

Answer

The area under the curve is \(\frac{1}{4}\).

Step 1: Divide the Interval into Subintervals

We are starting with the function \(y = x^3\), and the interval \([0, 1]\). First, divide this interval into \(n\) equal subintervals. The length of each subinterval is \(\Delta x = \frac{1-0}{n} = \frac{1}{n}\).

Step 2: Define the Right Endpoint for Each Subinterval

For each subinterval \(i\), the right endpoint is \(x_i = a + i\Delta x = 0 + i\frac{1}{n} = \frac{i}{n}\), where \(i = 1, 2, ..., n\).

Step 3: Calculate the Area of Each Rectangle

The height of each rectangle is determined by the function value at the right endpoint. So, the height of the \(i\)-th rectangle is \(f\left(\frac{i}{n}\right) = \left(\frac{i}{n}\right)^3\). The area of each rectangle is then \(f\left(\frac{i}{n}\right) \times \Delta x = \left(\frac{i}{n}\right)^3 \times \frac{1}{n} = \frac{i^3}{n^4}\).

Step 4: Sum the Areas of All Rectangles

Sum the areas of all rectangles from \(i = 1\) to \(n\):\[\sum_{i=1}^{n} \frac{i^3}{n^4}\ = \frac{1}{n^4}\sum_{i=1}^{n} i^3\]The sum \(\sum_{i=1}^{n} i^3\) is a known formula: \(\left(\frac{n(n+1)}{2}\right)^2\).

Step 5: Simplify the Summation

Substitute the sum formula into the area calculation:\[\frac{1}{n^4}\left(\frac{n(n+1)}{2}\right)^2 = \frac{1}{n^4}\left(\frac{n^2(n+1)^2}{4}\right) = \frac{(n+1)^2}{4n^2}\]

Step 6: Take the Limit as \(n\rightarrow \infty\)

Finally, take the limit of the sum as \(n\) approaches infinity:\[\lim_{n \to \infty} \frac{(n+1)^2}{4n^2} = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{4n^2} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4}\]Evaluating this as \(n\) goes to infinity, we get the result \(\frac{1}{4}\).

Step 7: Conclusion

The area under the curve \(y=x^3\) from \(0\) to \(1\) is \(\frac{1}{4}\). This is the area of the region under the curve using the limit process.

Key Concepts

Definite Integral

A definite integral is a core concept in integral calculus. It represents the accumulated quantity, such as an area under a curve, from one point to another. In the given exercise, we are calculating the area under the polynomial function \(y = x^3\) from \(x = 0\) to \(x = 1\). This is achieved through the process of integration over the interval \([0, 1]\).
The definite integral is denoted as \(\int_a^b f(x) \, dx\), where \(a\) and \(b\) are the bounds of integration. It calculates the net area under the curve, considering portions of the area above and below the x-axis.
To find this integral, we use a Riemann sum to approximate the area, and then refine our approximation by taking the limit as the number of subintervals approaches infinity. This approach accurately computes continuous areas by breaking them into finite, manageable pieces.

Riemann Sum

The Riemann sum is a method to approximate the total area under a curve over a specific interval by summing up areas of rectangles. In our problem, the Riemann sum estimates the area under \(y = x^3\) from \(x = 0\) to \(x = 1\).
We divided the interval into \(n\) subintervals, where each subinterval is of equal width \(\Delta x = \frac{1}{n}\). For each subinterval, we form rectangles whose heights are determined by the function value at the right endpoint \(x_i = \frac{i}{n}\).
Each rectangle's area is \(f\left(\frac{i}{n}\right) \times \Delta x\), and the total area is the sum of all such rectangles

• \(\sum_{i=1}^{n} \frac{i^3}{n^4}\).

The Riemann sum becomes more accurate as \(n\) increases, effectively providing a closer approximation to the area under the curve, and is crucial for understanding the concept of definite integrals.

Limit Process

The limit process helps transform the Riemann sum into a definite integral by shifting from a finite to an infinite number of subintervals. In the exercise, we evaluate the limit as \(n\) approaches infinity to fine-tune the area under \(y = x^3\).

• The summation of rectangle areas is expressed as \(\sum_{i=1}^{n} \frac{i^3}{n^4}\).
• This sum is simplified using the formula for the sum of cubes, leading to \(\frac{(n+1)^2}{4n^2}\).

By taking the limit \(\lim_{n \to \infty} \frac{n^2 + 2n + 1}{4n^2}\), you focus on the dominant terms as \(n\) grows.
As \(n\) becomes infinitely large, smaller terms vanish, and ultimately the calculation resolves to \(\frac{1}{4}\), representing the exact area.This mentality is fundamental in calculus, facilitating transitions from approximation to precision.

Polynomial Function

A polynomial function like \(y = x^3\) is an expression composed of variables and coefficients that involve only non-negative integer powers of the variable. In our exercise, \(x^3\) is a simple polynomial that exhibits the cubic growth of the function as it increases.
Polynomials are smooth and continuous, making them suitable for integral calculus, as they don't have abrupt changes in direction.
Key characteristics of a polynomial function include:

• Continuity, which ensures that the curve can be analyzed over any interval.
• Derivative properties, allowing for easy calculation of slopes and integrals.

These traits make polynomials ideal for learning fundamental calculus concepts, as they can be differentiated and integrated easily, providing clear insights into the behavior of functions over specified intervals.