Problem 57 - Calculus: Early Transcendentals

Question

Use the properties of integrals to verify the inequality without evaluating the integrals. \( \displaystyle 2 \le \int^1_{-1} \sqrt{1 + x^2} \,dx \le 2\sqrt{2} \)

Step-by-Step Solution

Answer

The inequality is verified by bounding the function and integrating these bounds over the interval.

Step 1: Understand the Function

First, let's identify the function inside the integral. We have \( f(x) = \sqrt{1 + x^2} \). This is a continuous and differentiable function on the interval \([-1, 1]\). It represents half of an elliptical integral calculation.

Step 2: Determine Bounds for the Function on the Interval

Evaluate the values of \( f(x) = \sqrt{1 + x^2} \) at the endpoints and within the interval. At \( x = -1 \) and \( x = 1 \), \( f(x) = \sqrt{2} \). At \( x = 0 \), \( f(x) = 1 \). Hence, \( 1 \leq f(x) \leq \sqrt{2} \) for \( x \) in \([-1, 1]\).

Step 3: Integrate the Minimum and Maximum Bounds

Integrate the bounds over \([-1, 1]\). For the lower bound: \( \int_{-1}^{1} 1 \, dx = 2 \). For the upper bound: \( \int_{-1}^{1} \sqrt{2} \, dx = 2\sqrt{2} \).

Step 4: Apply Integral Inequality

Using the properties of integrals, specifically that \( \int_{a}^{b} m \, dx \leq \int_{a}^{b} f(x) \, dx \leq \int_{a}^{b} M \, dx \) where \( m \leq f(x) \leq M \), apply this to our bounds: \( 2 \leq \int_{-1}^{1} \sqrt{1 + x^2} \, dx \leq 2\sqrt{2} \).

Step 5: Conclusion

Since the inequalities \( 2 \leq \int_{-1}^{1} \sqrt{1 + x^2} \, dx \leq 2\sqrt{2} \) hold based on the earlier evaluations, the verification using properties of integrals without computing the exact integral is complete.

Key Concepts

Properties of Integrals

The properties of integrals are essential tools in calculus, allowing us to work with integrals effectively without always computing them directly. One key property is the integral inequality, which states that if a function \(f(x)\) is bounded by two constants \(m\) and \(M\) over an interval \([a, b]\), then the integral of \(f(x)\) is also bounded by the integrals of these constants. This is expressed as:

• \( \int_a^b m \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b M \, dx \)

This property is powerful because it lets us establish bounds for an integral based on the function's behavior over an interval. In our exercise, this is used to conclude that the integral of \(\sqrt{1 + x^2}\) from \(-1\) to \(1\) lies between 2 and \(2\sqrt{2}\). We avoid computing the exact integral, saving time and complexity.

Continuous and Differentiable Functions

When we talk about continuous and differentiable functions, we are referring to functions that have no interruptions or breaks in their domain, and their derivatives exist at all points within that domain. For example, the function \(f(x) = \sqrt{1 + x^2}\) is both continuous and differentiable on the interval \([-1, 1]\).

A continuous function like \(f(x)\) ensures that there are no jumps or vertical asymptotes. This smoothness allows us to apply integration techniques and theorems confidently. A differentiable function, meanwhile, implies that we can calculate its rate of change, although in this particular context, our focus is more on its continuity which guarantees the integral bounds can be applied without disruptions.

Bounds Evaluation

Evaluating bounds is a crucial step in forming integral inequalities. To do this, we need to determine the minimum and maximum values that \(f(x) = \sqrt{1 + x^2}\) can take over the interval \([-1, 1]\).

• At \(x = 0\), \(f(x) = 1\).
• At \(x = 1\) and \(x = -1\), \(f(x) = \sqrt{2}\).

From these evaluations, we establish that \(1 \leq f(x) \leq \sqrt{2}\) for \(x\) within \([-1, 1]\). This gives us our lower bound of 1 and our upper bound of \(\sqrt{2}\). The integration of these bounds yields 2 and \(2\sqrt{2}\), respectively, which are then used to demonstrate the inequality without calculating the exact value of the integral.

Integral Approximation

Integral approximation methods, while not directly applied in this exercise, play a significant role in approximating difficult integrals. Understanding this helps appreciate how bounds evaluation serves as a basic approximation.

Methods such as Riemann sums, trapezoidal rule, and Simpson’s rule provide ways to estimate the value of an integral when its exact solution is complex to find. By approximating the area under a curve, these techniques give insights into the integral's magnitude and range. In our problem, instead of using these methods directly, we use the calculated bounds to approximate and verify that our integral falls within a specified range, which effectively gives us an estimate of its true value.