Problem 58 - Calculus

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 \(n \rightarrow \infty\). (See the example for \(y=x^{2}\) in the text.) \(y=x^{3}+x ; a=0, b=1\)

Step-by-Step Solution

Answer

The area is \( \frac{3}{4} \).

Step 1: Define the Function and Interval

We are given the function \( y = x^3 + x \) and the interval \([0, 1]\). We need to find the area under this curve over the interval \([0, 1]\).

Step 2: Divide the Interval into n Subintervals

Divide the interval \([0, 1]\) into \( n \) equal subintervals. The width of each subinterval, \( \Delta x \), is given by \( \Delta x = \frac{1-0}{n} = \frac{1}{n} \).

Step 3: Evaluate the Midpoint of Each Subinterval

For each subinterval, find the midpoint \( x_i = \frac{i-0.5}{n} \) where \( i = 1, 2, \, \ldots \, , n \).

Step 4: Calculate the Function Value at Each Midpoint

Calculate \( f(x_i) = (x_i)^3 + x_i \) for each midpoint. Substituting \( x_i \) gives \( f\left(\frac{i-0.5}{n}\right) = \left(\frac{i-0.5}{n}\right)^3 + \frac{i-0.5}{n} \).

Step 5: Sum the Areas of Rectangles

Form the sum of the areas of the rectangles using the heights at the midpoints. The area of the rectangle for each \( x_i \) is \( f(x_i) \Delta x \) thus the total estimated area is given by the Riemann sum: \[ S_n = \sum_{i=1}^{n} \left(\left(\frac{i-0.5}{n}\right)^3 + \frac{i-0.5}{n}\right) \cdot \frac{1}{n} \].

Step 6: Take the Limit of the Riemann Sum as n Approaches Infinity

The definite integral gives the area under the curve and is the limit of the Riemann sum as \( n \to \infty \): \[ \lim_{n \to \infty} S_n = \int_{0}^{1} (x^3 + x) \, dx \].

Step 7: Compute the Definite Integral

Using the power rule of integration, integrate \( \int (x^3 + x) \, dx \) from 0 to 1. This gives: \[ \int_{0}^{1} (x^3 + x) \, dx = \left. \left(\frac{x^4}{4} + \frac{x^2}{2}\right) \right|_0^1 = \left(\frac{1^4}{4} + \frac{1^2}{2}\right) - \left(\frac{0^4}{4} + \frac{0^2}{2}\right) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}. \]

Step 8: Conclusion

The area under the curve \( y = x^3 + x \) from \( a = 0 \) to \( b = 1 \) is \( \frac{3}{4} \).

Key Concepts

Understanding Riemann Sum

A Riemann sum is an essential tool used to approximate the area under a curve. This concept helps us move from a rough estimate to an accurate measurement. Imagine wanting to calculate the area under the curve of a function, such as the given function \( y = x^3 + x \), from \( a = 0 \) to \( b = 1 \).
The interval \([0, 1]\) is divided into \( n \) smaller subintervals. These subintervals have equal widths \( \Delta x = \frac{1}{n} \), allowing you to systematically explore each segment by creating rectangles across the interval.
The height of these rectangles corresponds to the function value at specific points within each subinterval, often at midpoints, like \( x_i = \frac{i-0.5}{n} \). You then sum up the area of all these rectangles. The formula becomes:

• Area of rectangle = height \( \times \) width = \( f(x_i) \Delta x \)
• Riemann sum \( S_n = \sum_{i=1}^{n} f(x_i) \Delta x \)

The beauty of Riemann sums is revealed when \( n \rightarrow \infty \), transitioning from approximation to precision as it approaches the true area under the curve.

Applying the Power Rule of Integration

Integration is the process of finding the area under the curve by precisely adding up infinitely small widths, a stark transition from Riemann sums to definite integrals. The power rule of integration makes solving integrals easier because it provides a straightforward approach for integrating polynomial functions.
If you have a term, like \( x^n \), the power rule of integration states:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

To apply this to the given function \( y = x^3 + x \), you integrate both terms separately over \([0, 1]\):

• \( \int x^3 \, dx = \frac{x^4}{4} + C \)
• \( \int x \, dx = \frac{x^2}{2} + C \)

The definite integral then evaluates these from \( 0 \) to \( 1 \), using the fundamental theorem of calculus to find the sum of the areas, ensuring accurate results.

Calculating the Area Under the Curve

The area under the curve represents the total value accumulated between the curve and the x-axis over a specific interval. In this context, it is the result of the definite integral of the given function \( y = x^3 + x \) from \( 0 \) to \( 1 \).
To find this, you follow the integration process and employ the limits of the integral:

• Set up the integral \( \int_{0}^{1} (x^3 + x) \, dx \).
• Use the power rule to integrate each term, resulting in \( \frac{x^4}{4} + \frac{x^2}{2} \).
• Evaluate this from \( x = 0 \) to \( x = 1 \).

Finally, substituting the bounds into the antiderivative gives \( \left( \frac{1}{4} + \frac{1}{2} \right) - \left( 0 \right) = \frac{3}{4} \).
This reflects the total area under the curve, confirming our solution: the calculation becomes an elegant process, bridging numerical estimation and symbolic integration to solve real-world problems.