/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Let \(f\) and \(g\) be two conti... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f\) and \(g\) be two continuous functions such that \(0 \leq m_{1} \leq f(x) \leq M_{1}\) for any \(x \in[a, b]\) and \(0 \leq m_{2} \leq g(y) \leq M_{2}\) for any \(y \in[c, d] .\) Show that the following inequality is true: \(m_{1} m_{2}(b-a)(c-d) \leq \int_{a}^{b} \int_{c}^{d} f(x) g(y) d y d x \leq M_{1} M_{2}(b-a)(c-d)\)

Short Answer

Expert verified
The inequality holds by bounding \(f(x)g(y)\) as shown through integral evaluation and applying function bounds.

Step by step solution

01

Express the Double Integral

The inequality involves a double integral of the product of two functions, \(f(x)\) and \(g(y)\), over the region \([a, b] \times [c, d]\). We write this as:\[ I = \int_{a}^{b} \int_{c}^{d} f(x) g(y) \, dy \, dx. \]
02

Examine the Bounds of f(x)g(y)

Given that \(m_1 \leq f(x) \leq M_1\) and \(m_2 \leq g(y) \leq M_2\), the product \(f(x)g(y)\) is bounded by:\[ m_1 m_2 \leq f(x)g(y) \leq M_1 M_2. \]
03

Simplify the Double Integral Using Bounds

Applying the bounds from Step 2, we can bound the double integral:\[ \int_{a}^{b} \int_{c}^{d} m_1 m_2 \, dy \, dx \leq \int_{a}^{b} \int_{c}^{d} f(x) g(y) \, dy \, dx \leq \int_{a}^{b} \int_{c}^{d} M_1 M_2 \, dy \, dx. \]
04

Evaluate the Simplified Integral Limits

Integrate the simplified bounds:- The lower bound is: \[ \int_{a}^{b} \int_{c}^{d} m_1 m_2 \, dy \, dx = m_1 m_2 \cdot (d-c)(b-a). \]- The upper bound is: \[ \int_{a}^{b} \int_{c}^{d} M_1 M_2 \, dy \, dx = M_1 M_2 \cdot (d-c)(b-a). \]
05

Establish the Final Inequality

Substitute the evaluated integrals back into the inequality from Step 3, leading to:\[ m_1 m_2 (b-a)(d-c) \leq I \leq M_1 M_2 (b-a)(d-c), \]this shows that the double integral lies between these computed bounds.

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Key Concepts

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

Continuous Functions
Continuous functions are a fundamental concept in calculus and analysis. They are functions that do not have any abrupt changes or breaks. More formally, a function is continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. Continuous functions have two critical properties that are useful in double integrals:

- They can be integrated smoothly over any closed interval.
- Their behavior is predictable within the integration bounds. In the context of the exercise, both functions \( f(x) \) and \( g(y) \) are continuous within their respective intervals \([a, b]\) and \([c, d]\). This continuity ensures that we can compute their product within these intervals using double integration. Since there are no interruptions or irregularities, the integration process is straightforward, ensuring accurate results. Continuous functions are essential in maintaining the integrity of integration processes.
Inequalities in Integrals
Inequalities in integrals help bound the values of an integral from above and below, providing a range where the actual value lies. In this exercise, we use inequalities to set limits on the double integral of the function products \( f(x) g(y) \).

- If \(m_1 \leq f(x) \leq M_1\) and \(m_2 \leq g(y) \leq M_2\), it follows that the product \( m_1 m_2 \leq f(x)g(y) \leq M_1 M_2 \).
- These product bounds can then be integrated over the set boundaries, \([a,b]\) and \([c,d]\), to produce bounds for the entire double integral.By integrating these bounds, we achieve:
\[ m_1 m_2 (b-a)(d-c) \leq I \leq M_1 M_2 (b-a)(d-c) \]
where \( I \) represents the double integral. This step is crucial for verifying that the computed integral will always lie within these constraints, ensuring accuracy and reliability in results.
Integration Bounds
Integration bounds are the limits between which an integral is evaluated. Specifying correct bounds is critical for accurate integration. In this exercise, the integration is performed over a rectangular region defined by the bounds \([a, b] \) for \( x \) and \([c, d] \) for \( y \).

- The bounds supply the parameters in which we calculate the double integral of the product function \( f(x)g(y) \).
- These bounds translate into the limits of the outer and inner integral calculations, indicating the region we're dealing with.In mathematical terms, the integral is evaluated as:
\[ \int_{a}^{b} \int_{c}^{d} f(x)g(y) \, dy \, dx \]
This means that for every value of \( x \), \( y \) varies from \( c \) to \( d \), and for every value of \( y \), \( x \) varies from \( a \) to \( b \). Understanding and setting these bounds correctly is essential for properly defining the area over which the function is integrated.
Product of Functions
The product of functions is a concept key to this exercise since we're dealing with the double integral of such a product, \( f(x) g(y) \), where each function depends on a different variable.

- The function \( f(x) \) depends on \( x \) while \( g(y) \) depends on \( y \).
- The product \( f(x) g(y) \) is a new function spanning the variables \( x \) and \( y \).This multiplication is crucial as it affects the integration process:
- The resulting function \( f(x) g(y) \) is bound by the inequalities derived from the maximum and minimum values of \( f(x) \) and \( g(y) \).
- Properly handling the product of functions is vital when establishing the bounding inequalities used for calculating double integrals.By first bounding each function individually, we can confidently integrate their product knowing it remains within fixed limits over the integrated area. This ensures that calculations perform smoothly and within the expected framework.

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Most popular questions from this chapter

The following problems examine Mount Holly in the state of Michigan. Mount Holly is a landfill that was converted into a ski resort. The shape of Mount Holly can be approximated by a right circular cone of height 1100 ft and radius 6000 \(\mathrm{ft.}\) In reality, it is very likely that the trash at the bottom of Mount Holly has become more compacted with all the weight of the above trash. Consider a density function with respect to height: the density at the top of the mountain is still density 400 \(\mathrm{lb} / \mathrm{ft}^{3}\) and the density increases. Every 100 feet deeper, the density doubles. What is the total weight of Mount Holly?

The power emitted by an antenna has a power density per unit volume given in spherical coordinates by $$ p(\rho, \theta, \varphi)=\frac{P_{0}}{\rho^{2}} \cos ^{2} \theta \sin ^{4} \varphi $$ where \(P_{0}\) is a constant with units in watts. The total power within a sphere \(B\) of radius \(r\) meters is defined as \(P=\iiint_{B} p(\rho, \theta, \varphi) d V .\) Find the total power \(P .\)

Show that the volume of a regular right hexagonal pyramid of edge length \(a\) is \(\frac{a^{3} \sqrt{3}}{2}\) by using triple integrals.

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For the following problems, find the specified area or volume. The volume of the intersection between two spheres of radius \(1,\) the top whose center is \((0,0,0.25)\) and the bottom, which is centered at \((0,0,0)\) For the following problems, find the center of mass of the region.

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