Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \(1 \leq \int_{0}^{1} \sqrt{1+x^{4}} d x \leq \frac{6}{5}\). Hint: Explain why \(1 \leq \sqrt{1+x^{4}} \leq 1+x^{4}\) for \(x\) in the closed interval \([0,1]\); then use the Comparison Property (Theorem B) and the result of Problem 39d.

Step-by-Step Solution

Answer

The inequality holds: \( 1 \leq \int_{0}^{1} \sqrt{1+x^{4}} \, dx \leq \frac{6}{5} \).

Step 1: Analyze the function

Let's first analyze the inequality for the function inside the integral: \( \sqrt{1+x^4} \). Notice that \( x^4 \) is always non-negative for \( x \) in \([0, 1]\). Therefore, \( \sqrt{1+x^4} \) is always greater than or equal to 1, since \( \sqrt{1} = 1 \). Also, \( \sqrt{1+x^4} \) should always be less than or equal to \( 1 + x^4 \) because the square root function grows slower than the linear term as \( x \) increases from 0 to 1.

Step 2: Establish inequalities for the integral

We have the inequalities \( 1 \leq \sqrt{1+x^4} \leq 1+x^4 \) for \(x \) in the interval \([0,1]\). By using these bounds, we establish the inequalities for the integrals: \[ \int_{0}^{1} 1 \, dx \leq \int_{0}^{1} \sqrt{1+x^4} \, dx \leq \int_{0}^{1} (1+x^4) \, dx. \]

Step 3: Calculate the lower bound integral

Calculate the integral of 1 from 0 to 1: \( \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \). This gives us the lower bound for the original integral.

Step 4: Calculate the upper bound integral

Calculate the integral of \(1 + x^4\) from 0 to 1: \[ \int_{0}^{1} (1 + x^4) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} x^4 \, dx. \] The first integral is 1. Now compute \( \int_{0}^{1} x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5} - 0 = \frac{1}{5} \). Thus, the second integral equals \(1 + \frac{1}{5} = \frac{6}{5} \).

Step 5: Conclude with the Comparison Property

Using the Comparison Property (a theorem that allows comparisons when one function bounds another over an interval), the inequality holds: \[ 1 \leq \int_{0}^{1} \sqrt{1+x^4} \, dx \leq \frac{6}{5}. \] This conclusion follows because we have compared the integral with two bounds we managed to compute: 1 as the lower bound and \(\frac{6}{5}\) as the upper bound.

Key Concepts

Integration Techniques

Integration Techniques are crucial when solving calculus problems involving finding the area under a curve, like the problem at hand with the integral of \( \sqrt{1+x^4} \) from 0 to 1.
There are various techniques that can make finding integrals easier depending on the function you are working with.
In this particular example, we recognize two helpful techniques:

• Simplifying the function: By observing inequalities, we notice that \(1 \leq \sqrt{1+x^4} \leq 1+x^4\), which already provides a pathway to estimate the integral.
• Splitting functions: When dealing with \( \int (1 + x^4) \, dx \), we can split it into \( \int 1 \, dx \) and \( \int x^4 \, dx \) which are easier to solve.

The beauty of integration techniques lies in breaking down the problem into more manageable parts.

Comparison Theorem

The Comparison Theorem is a powerful tool in calculus for determining the bounds of an integral by comparing it to more straightforward functions. In our exercise, it is used to establish the inequality:\[1 \leq \int_{0}^{1} \sqrt{1+x^4} \, dx \leq \frac{6}{5}.\]
This theorem involves comparing the function \( \sqrt{1+x^4} \) with the simpler functions 1 and \(1 + x^4\):
- A function \( f(x) \) is bound by \( g(x) \) and \( h(x) \) if \( g(x) \leq f(x) \leq h(x) \) over an interval.- By integrating each at the bounds, we find the constraints for the more complex integral.
In this problem, because \( \sqrt{1+x^4} \) always stays between 1 and \(1 + x^4\), the integral of \( \sqrt{1+x^4} \) also falls between the integrals of these simpler functions.

Calculus Problem Solving

Calculus Problem Solving involves a series of logical steps that help in navigating complex mathematical problems.
Solving the given exercise requires:

• Understanding the behavior of the function \( \sqrt{1+x^4} \) over the interval [0, 1].
• Setting up inequalities, \( 1 \leq \sqrt{1+x^4} \leq 1+x^4 \), which suggest a range for the integral.
• Applying known theorems like the Comparison Theorem which justifies the bounds.
• Executing straightforward integrations for the bounds which, in this case, are constants and polynomial functions.

This structured approach demonstrates the essence of calculus problem solving—breaking down a problem into understandable parts and integrating formal mathematical theory with practical computation.

Definite Integrals

Definite Integrals are essential in calculus as they help to calculate the exact area under a curve over an interval. In our exercise, we compute:

• \( \int_{0}^{1} 1 \, dx = 1 \)
• \( \int_{0}^{1} (1 + x^4) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} x^4 \, dx = 1 + \frac{1}{5} = \frac{6}{5} \)

Calculating these integrals involves finding an antiderivative or primitive of the function and then applying the evaluation theorem \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f(x) \).
This exercise exemplifies how definite integrals yield not only specific numerical results but also support broader analytical conclusions about function behavior in calculus.