Problem 58 - 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 \(n \rightarrow \infty\). $$ y=x^{3}+x ; a=0, b=1 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to calculate the area under the curve described by the function \(y = x^3 + x\) over the interval \([0, 1]\). This involves integration, which calculates the area beneath the curve by considering an infinite number of infinitesimally small strips (rectangles) under the curve from \(x = 0\) to \(x = 1\).

Step 2: Set Up the Integral

To find the area under the curve, set up the definite integral of the function from 0 to 1. The integral to evaluate is \(\int_{0}^{1} (x^3 + x)\, dx\).

Step 3: Integrate the Function

Find the antiderivative of the function \(x^3 + x\). The antiderivative of \(x^3\) is \(\frac{x^4}{4}\), and the antiderivative of \(x\) is \(\frac{x^2}{2}\). Therefore, the antiderivative of \(x^3 + x\) is \(\frac{x^4}{4} + \frac{x^2}{2}\).

Step 4: Evaluate the Definite Integral

Evaluate the antiderivative from 0 to 1. Substitute \(x = 1\) into the antiderivative: \(\frac{1^4}{4} + \frac{1^2}{2} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}\). Now substitute \(x = 0\), which gives \(\frac{0^4}{4} + \frac{0^2}{2} = 0\). The area under the curve is \(\frac{3}{4} - 0 = \frac{3}{4}\).

Key Concepts

Definite Integral

When working with problems involving finding areas under curves, a definite integral is one of the main tools used. A definite integral gives the accumulated area between a given function and the x-axis across a specific interval. In this case, we want to find the area under the curve of the function \( y = x^3 + x \) from \( x = 0 \) to \( x = 1 \). This involves calculating the integral of the function over this interval.

The process requires setting up an integral like \( \int_{0}^{1} (x^3 + x) \, dx \). Here, \( a = 0 \) and \( b = 1 \) represent the bounds of integration. The limits tell us the start and end point along the x-axis for which we're computing the area. During integration, small, infinitely tiny sections under the curve are summed up, resulting in the total area beneath the curve.

Key features of definite integrals include:

• The integrand: This is the function being integrated, in our case, \( x^3 + x \).
• The limits of integration: The values \( a \) and \( b \), dictating the interval over which the integration is performed.
• Riemann sums: An approach used to approximate the area by summing rectangular section areas under the curve, as \( n \rightarrow \infty \).

Antiderivative

Antiderivatives are the backbone of solving definite integrals. They essentially reverse the operation of differentiation by finding a function whose derivative matches the original function you're working with. For the definite integral we set up, we need to determine the antiderivative of \( y = x^3 + x \).

To find the antiderivative, identify simpler integrations:

• The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \)
• The antiderivative of \( x \) is \( \frac{x^2}{2} \)

Therefore, the antiderivative of \( x^3 + x \) is given by combining these: \( \frac{x^4}{4} + \frac{x^2}{2} \). This new function is used to evaluate the definite integral, providing the total area beneath the curve over the given interval.

Evaluating at boundaries:

• Substitute the upper limit (1) into the antiderivative
• Subtract the result of substituting the lower limit (0)

This calculation directly gives the area under the curve.

Area Under a Curve

The concept of finding the area under a curve is crucial in calculus, as it gives us a visual quantification of this accumulated space. This is especially important in real-world applications where you're interested in things like accumulated distance or total production over a time period.

When solving our problem, the curve described by \( f(x) = x^3 + x \) is analyzed between 0 and 1. The area under this curve from 0 to 1 is precisely the result of the integral, which we found to be \( \frac{3}{4} \). This means that between \( x = 0 \) and \( x = 1 \), the total enclosed area between the function and the x-axis is \( \frac{3}{4} \, \text{square units} \).

Other key points of note include:

• By letting \( n \rightarrow \infty \)—increasingly smaller rectangles—the approximation becomes exact.
• The concept relies on a fundamental theorem of calculus: linking antiderivatives to evaluate definite integrals.

Understanding the area under a curve is foundational, shaping many problem-solving strategies in both theoretical and applied mathematics.