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Chapter 4: Problem 45

Find the value(s) of \(c\) guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. $$ f(x)=2 \sec ^{2} x, \quad[-\pi / 4, \pi / 4] $$

Short Answer

Expert verified
The value of \( c \) that satisfies the Mean Value Theorem for Integrals for the given function and interval is \( c = 0 \).

Step by step solution

01

Calculate \( \int_{a}^{b} f(x) dx \)

Begin by calculating the integral of \( f(x) \) from \( a \) to \( b \). The function \( f(x) \) is given as \( 2 \sec^2(x) \), so the integral from \( -\pi/4 \) to \( \pi/4 \) can be written as \( \int_{-\pi/4}^{\pi/4} 2 \sec^2(x) dx \). Integrating this gives \( F(b) - F(a) = 2 \tan(\pi/4) - 2 \tan(-\pi/4) = 2(1) - 2(-1) = 4 \).
02

Setup Mean Value Theorem for Integrals equation

For the Mean Value Theorem for Integrals to hold true, \[ f(c)(b - a) = \int_{a}^{b} f(x) dx \]. We'll replace \( \int_{a}^{b} f(x) dx \) with the result from the first step (which was 4) and \( (b - a) \) with \( \pi/4 - (-\pi/4) \) or \( \pi/2 \). Therefore: \( 2 \sec^2(c)(\pi/2) = 4 \].
03

Solve for \( c \)

Now solve the equation to find the value(s) of \( c \). Dividing both sides of the equation by \( \pi \), gives \( 2 \sec^2(c) = 4/\pi \). Solving for \( \sec^2(c) \), gives \( \sec^2(c) = 2/\pi \). Taking the square root of both sides gives \( \sec(c) = \sqrt{2/\pi} \). Since \( \sec (c) \) is \( 1 /\cos(c) \), we can find \( \cos(c) = \sqrt{\pi/2} \). Lastly, since \( c \) should be within the interval \( [-\pi/4, \pi/4] \), the only valid solution is \( c = 0 \) as long as \( \pi/2 > 1 \), which is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrals
An integral can be understood as the area under a curve. It gives us a way to add up infinitely small slices to find a whole. In the context of the exercise, we want to find the area under the curve defined by the function \( f(x) = 2 \sec^2(x) \) over a interval, specifically from \(-\pi/4\) to \(\pi/4\).

To calculate this integral, we use the antiderivative of \( f(x) \). Here, since \( \sec^2(x) \) is the derivative of \( \tan(x) \), we have \( \int 2 \sec^2(x) \, dx = 2 \tan(x) \). Evaluate this from \(-\pi/4\) to \(\pi/4\), meaning we calculate the difference \( 2 \tan(\pi/4) - 2 \tan(-\pi/4) \).

Remember:
  • The integral helps in determining complete quantities like area.
  • Understanding the integral's geometric interpretation eases problem solving.
  • Ensure to evaluate at both the upper and lower bounds, not just the upper one.
Secant Function
The secant function \( \sec(x) \) is a trigonometric function that is the reciprocal of the cosine function, \( \sec(x) = 1/\cos(x) \). It describes how the cosine function's behavior influences other trigonometric identities and equations.

In this exercise, we were given \( \sec^2(x) \) which is often encountered in calculus, particularly in the context of integrals and derivatives. When \( \sec(x) \) appears squared, it strongly suggests the involvement of tangent due to the identity \( \sec^2(x) = 1 + \tan^2(x) \).

Why is this important?
  • Secant becomes very large near points where cosine is zero, watch out for vertical asymptotes.
  • It is directly related to the slopes (derivatives) of tangent functions, useful for solving integrals.
  • Knowing the reciprocal relationship is key to solving equations.
Trigonometric Integration
Trigonometric integration involves solving integrals with trigonometric functions. It's crucial for finding areas and solving equations involving oscillatory behavior.

For functions like \( \sec^2(x) \), trigonometric identities are often the simplest path to a solution. These identities help transform functions into integrable forms. In our exercise, we used the fact that \( \sec^2(x) \) is the derivative of \( \tan(x) \) to quickly integrate \( 2 \sec^2(x) \).

Benefits of understanding trigonometric integration:
  • Simplifies complex integrals involving angles.
  • Helps solve tasks related to periodic motions or cycles.
  • Useful in physics, engineering, and other applied fields.
Calculus
Calculus is the branch of mathematics focusing on rates of change and accumulation of quantities. It provides the tools needed to understand motion, growth, and areas under curves.

Key concepts in calculus include differential and integral calculus. Differential calculus deals with instantaneous rates of change, while integral calculus concerns cumulative quantities. In this exercise, the Mean Value Theorem for Integrals, a fundamental calculus theorem, helps find specific points within an interval that represent an average value.

Remember, calculus allows us to:
  • Model physical systems through derivatives and integrals.
  • Find solutions to real-world problems involving growth and decay.
  • Develop a deeper understanding of continuous change.

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Most popular questions from this chapter

Find the indefinite integral and check the result by differentiation. $$\int\left(\sec ^{2} \theta-\sin \theta\right) d \theta$$

The table shows the velocities (in miles per hour) of two cars on an entrance ramp to an interstate highway. The time \(t\) is in seconds. $$ \begin{array}{|l|c|c|c|c|c|c|c|} \hline t & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline v_{1} & 0 & 2.5 & 7 & 16 & 29 & 45 & 65 \\ \hline v_{2} & 0 & 21 & 38 & 51 & 60 & 64 & 65 \\ \hline \end{array} $$ (a) Rewrite the table converting miles per hour to feet per second. (b) Use the regression capabilities of a graphing utility to find quadratic models for the data in part (a). (c) Approximate the distance traveled by each car during the 30 seconds. Explain the difference in the distances.

A differential equation, a point, and a slope field are given. A slope field (or direction field) consists of line segments with slopes given by the differential equation. These line segments give a visual perspective of the slopes of the solutions of the differential equation. (a) Sketch two approximate solutions of the differential equation on the slope field, one of which passes through the indicated point. (To print an enlarged copy of the graph, select the MathGraph button.) (b) Use integration to find the particular solution of the differential equation and use a graphing utility to graph the solution. Compare the result with the sketches in part (a). $$ \frac{d y}{d x}=x^{2}-1, \quad(-1,3) $$

Slope Fields (a) use a graphing utility to graph a slope field for the differential equation, (b) use integration and the given point to find the particular solution of the differential equation, and (c) graph the solution and the slope field in the same viewing window. $$\frac{d y}{d x}=2 \sqrt{x},(4,12)$$

Find \(F^{\prime}(x)\). $$ F(x)=\int_{0}^{\sin x} \sqrt{t} d t $$
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