Problem 77 - Calculus for Biology and Medicine

Question

Verify each inequality without evaluating the integrals. $$ 0 \leq \int_{0}^{4} \sqrt{x} d x \leq 8 $$

Step-by-Step Solution

Answer

The inequality \(0 \leq \int_{0}^{4} \sqrt{x} \, dx \leq 8\) is valid as \(\sqrt{x}\) is non-negative and \(\int_{0}^{4} \sqrt{x} \, dx\) is bounded by 8.

Step 1: Understand the Exercise

We need to verify that \(0 \leq \int_{0}^{4} \sqrt{x} \, dx \leq 8\) without solving the integral directly. This involves comparing the integral to known bounds.

Step 2: Establish the Lower Bound

The function \(\sqrt{x}\) is non-negative for all \(x\) in the interval \([0, 4]\) because the square root of a non-negative number is always non-negative. Thus, the integral of \(\sqrt{x}\) from \(0\) to \(4\) is also non-negative. This confirms the lower bound of \(0 \leq \int_{0}^{4} \sqrt{x} \, dx\).

Step 3: Estimate the Upper Bound

Evaluate the maximum value of the function \(\sqrt{x}\) over the interval \([0, 4]\). Since \(\sqrt{x}\) is an increasing function, its maximum value is at the right endpoint, \(x = 4\). Here, \(\sqrt{4} = 2\).

Step 4: Calculate the Upper Bound Using Rectangle Approximation

Using the rectangle method, the maximum height of the rectangle is \(2\) (from Step 3) over the interval length \(4\). Therefore, the area is \((4) \times (2) = 8\). This proves that \(\int_{0}^{4} \sqrt{x} \, dx \leq 8\).

Step 5: Combine the Results

Since both conditions \(0 \leq \int_{0}^{4} \sqrt{x} \, dx\) and \(\int_{0}^{4} \sqrt{x} \, dx \leq 8\) are satisfied, we verify the compound inequality \(0 \leq \int_{0}^{4} \sqrt{x} \, dx \leq 8\) without evaluating the integral.

Key Concepts

Definite Integrals

Definite integrals play a crucial role in capturing the area under a curve within a specified interval. Imagine the graph of a function; the definite integral from a point \(a\) to \(b\) essentially measures the total area enclosed by the graph, the x-axis, and the vertical lines at \(x = a\) and \(x = b\).

In the exercise, the integral \(\int_{0}^{4} \sqrt{x} \, dx\) represents the area under the square root function, \(\sqrt{x}\), over the interval \([0, 4]\). This particular integral is interesting because it describes a common area computation, yet we are tasked with verifying an inequality for this area without actually calculating the integral.

• The integral bounds \(0\) to \(4\) tell us the section of the graph we are considering.
• Definite integrals are always associated with a real number, representing that specific area.

This concept is foundational in calculus, offering insights into real-world phenomena such as accumulated quantities over time.

Upper and Lower Bounds

Establishing upper and lower bounds is a method of determining the range within which an integral lies. It's a way of narrowing down the potential value of a definite integral without calculating it completely.

In our exercise, the inequality \(0 \leq \int_{0}^{4} \sqrt{x} \, dx \leq 8\) sets the boundaries for our integral over the interval \([0, 4]\). Here's why these bounds hold:

• Lower Bound: The function \(\sqrt{x}\) is always non-negative when \(x\) is positive or zero, so the integral cannot be less than zero.
• Upper Bound: To estimate an upper bound, we consider the largest value \(\sqrt{x}\) attains on \([0, 4]\), which is 2. Multiplying this maximum by the interval width 4, gives an area less than or equal to 8.

This logical bounding technique allows us to affirm the inequality without needing to evaluate the integral directly. It's a powerful tool for judging integrals in a broader context.

Square Root Function

The square root function, denoted as \(f(x) = \sqrt{x}\), is quite distinctive due to its properties and real-world applications. Here's a quick dive into some essentials:

• The function \(\sqrt{x}\) is only defined for non-negative values of \(x\).
• It is an increasing function, meaning as \(x\) increases, \(\sqrt{x}\) increases as well.
• The curve starts at the origin (0,0) and rises slowly, showcasing a plateau effect as \(x\) becomes very large.

For the interval \([0, 4]\), \(\sqrt{x}\) maintains a gentle slope that increases continuously. Calculus explores these properties further, often analyzing behavior like this to generate insights into geometric transformations, signal processing, and more. Recognizing \(\sqrt{x}\)'s behavior helps us understand the bounds and behavior of related integrals effectively.