Problem 66 - Calculus Single Variable

Question

A definite integral is given. Do not attempt to calculate its value \(V .\) Instead, find the extreme values of the integrand on the interval of integration, and use these extreme values together with the inequalities of line (5.3.5) to obtain numbers \(A\) and \(B\) such that \(A \leq V \leq B\). $$ \int_{0}^{3 / 2}(\sqrt{x}-\sin (x)) d x $$

Step-by-Step Solution

Answer

Integrand is \(\sqrt{x} - \sin(x)\) and extreme values determine bounds \(A\) and \(B\) for the integral.

Step 1: Identify the Integrand

The integrand is given as \(f(x) = \sqrt{x} - \sin(x)\). Our task is to find the extreme values of this function over the interval \([0, \frac{3}{2}]\).

Step 2: Find the Derivative

Calculate the derivative of the function to find its critical points: \(f'(x) = \frac{1}{2\sqrt{x}} - \cos(x)\).

Step 3: Solve for Critical Points

Set the derivative equal to zero and solve for \(x\): \(\frac{1}{2\sqrt{x}} = \cos(x)\). This equation is generally solved using numerical methods or by examining values, as there is no simple algebraic solution.

Step 4: Evaluate at Endpoints and Critical Points

Compute values of \(f(x)\) at the endpoints \(x = 0\), \(x = \frac{3}{2}\), and any critical points found in Step 3. For \(x = 0\): \(f(0) = 0 - 0 = 0\). For \(x = \frac{3}{2}\): \(f\left(\frac{3}{2}\right) = \sqrt{\frac{3}{2}} - \sin\left(\frac{3}{2}\right)\). Critical points need to be checked numerically.

Step 5: Apply the Extreme Value Theorem

From Step 4, determine the minimum (\(m\)) and maximum (\(M\)) values of \(f(x)\) on the interval \([0, \frac{3}{2}]\).

Step 6: Use the Integral Bounds

Using the inequality from calculus \(m(b-a) \leq \int_a^b f(x) \, dx \leq M(b-a)\), and substituting the interval \([0, \frac{3}{2}]\), we get: \[ m \cdot \frac{3}{2} \leq \int_0^{\frac{3}{2}} f(x) \, dx \leq M \cdot \frac{3}{2} \].

Step 7: Calculate and Assign A and B

If the extreme values are evaluated to be \( m \) and \( M \), compute \( A = m \cdot \frac{3}{2} \) and \( B = M \cdot \frac{3}{2} \). This provides the bounds \( A \leq V \leq B \).

Key Concepts

Extreme Values

The concept of extreme values is fundamental in finding the bounds for a definite integral. To determine these values, we analyze the integrand, which is the function we are integrating, over our specified interval. Extreme values refer to the minimum and maximum values that the integrand can take within this range.

• First, we calculate the derivative of the function. This helps us find points where the function could have local maxima or minima, called critical points.
• Once we locate these critical points, evaluating the function at these points (along with the endpoints of the interval) allows us to determine the smallest and largest possible outputs of the function.
• These outputs, the extreme values, help set bounds on the range of the definite integral.

Recognizing and determining these extreme values ensures accurate approximation of integrals without direct computation.

Integrand

In an integral, the integrand is the function that we are integrating over a specific interval. It is crucial to identify this function accurately, as it forms the basis of the calculations and concepts involved in solving an integral problem.

• In our exercise, the integrand is given as \(f(x) = \sqrt{x} - \sin(x)\).
• The evaluation of this function across the interval \([0, \frac{3}{2}]\) reveals insights into its behavior and helps find the extreme values.
• For applications like setting boundaries for the integral, understanding the nature of this function is essential.

Paying attention to the integrand can reveal characteristics, such as continuity and behavior, which significantly impact the solution of the integral.

Derivative

The derivative plays a vital role in calculus, especially when finding points of interest like critical points. For our integrand, the derivative helps identify where extreme values occur.

• By computing the derivative \(f'(x) = \frac{1}{2\sqrt{x}} - \cos(x)\), we can explore where this function increases or decreases.
• Setting the derivative equal to zero, \(\frac{1}{2\sqrt{x}} = \cos(x)\), highlights potential locations for minima or maxima.
• This equation may not always have straightforward solutions and sometimes requires numerical methods to solve.

Through the derivative, we determine regions of increasing or decreasing behavior, ultimately aiding in pinning down the function’s extreme values.

Endpoints

Examining the endpoints of an interval is a straightforward yet crucial step in finding extreme values for a function defined within that interval.

• In our interval \([0, \frac{3}{2}]\), endpoints are precisely at 0 and \(\frac{3}{2}\).
• These points are examined for their functional values to determine if they could potentially be the smallest or largest, i.e., extreme values.
• We calculate these explicitly: \(f(0) = 0 - 0 = 0\); and \(f(\frac{3}{2}) = \sqrt{\frac{3}{2}} - \sin(\frac{3}{2})\).

Checking the values at these endpoints and comparing them with any critical points found via the derivative helps ensure no potential maximum or minimum is overlooked. Integrating the concept of endpoints effectively allows for a comprehensive assessment of the function across its domain.