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Let \(M\) have continuous partials on a rectangle bounded by \(x=a, x=b, y=c\), and \(y=d\). Show that \(\int_{a}^{b} \frac{\partial M}{\partial x}(x, y) d x=M(b, y)-M(a, y) \quad\) for \(c \leq y \leq d\) and \(\int_{c}^{d} \frac{\partial M}{\partial y}(x, y) d y=M(x, d)-M(x, c) \quad\) for \(a \leq x \leq b\)

Short Answer

Expert verified
The integral property holds by the Fundamental Theorem of Calculus.

Step by step solution

01

Understand the Problem

We are given that the function \( M \) has continuous partial derivatives within a specified rectangle, and we need to confirm a specific integral property related to these partial derivatives.
02

Apply the Fundamental Theorem of Calculus for Part 1

For the integral \( \int_{a}^{b} \frac{\partial M}{\partial x}(x, y) \, dx \), where \( y \) is constant, use the Fundamental Theorem of Calculus. This theorem states that if a function has continuous derivatives, the integral of its derivative over an interval is equal to the net change over that interval. Therefore, \( \int_{a}^{b} \frac{\partial M}{\partial x}(x, y) \, dx = M(b, y) - M(a, y) \).
03

Verify the Conditions for Part 1

Ensure \( c \leq y \leq d \). Since \( M \) has continuous partial derivatives in this domain, the Fundamental Theorem of Calculus is applicable.
04

Apply the Fundamental Theorem of Calculus for Part 2

Consider the integral \( \int_{c}^{d} \frac{\partial M}{\partial y}(x, y) \, dy \), where \( x \) is constant. By the Fundamental Theorem of Calculus, this integral equals \( M(x, d) - M(x, c) \), since the partial derivative with respect to \( y \) is continuous.
05

Verify the Conditions for Part 2

Ensure \( a \leq x \leq b \). The continuity of the partial derivative \( \frac{\partial M}{\partial y} \) ensures this result is valid according to the theorem.

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Key Concepts

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

Continuous Partial Derivatives
In multivariable calculus, the concept of partial derivatives helps us understand the rate at which a function changes with respect to one variable while keeping others constant. When we talk about continuous partial derivatives, it means that these rates of change in all directions do not suddenly shift or break at any point within our domain.

For a function like \( M(x, y) \), having continuous partial derivatives \( \frac{\partial M}{\partial x} \) and \( \frac{\partial M}{\partial y} \) implies a smooth surface without any jumps or sharp edges. This continuity is crucial because it allows us to reliably use calculus operations, like integration, ensuring the solutions are predictable and accurate.
  • The function's behavior is smooth and consistent.
  • Allows for the application of important theorems, such as the Fundamental Theorem of Calculus.
Rectangular Domain
The rectangular domain is an interval within which we explore the behavior of our function. It's defined by the boundaries \( x=a \), \( x=b \), \( y=c \), and \( y=d \). Essentially, this creates a rectangle in the plane where we focus our analysis.

In the context of this exercise, the rectangle ensures that our function \( M(x, y) \) is examined over a well-determined section of two-dimensional space. It's important when verifying conditions like whether the partial derivatives are continuous across this area.
  • Provides a clear and confined area of study.
  • Ensures uniformity in analyzing function behavior.
  • Helps establish boundaries for integration purposes.
Line Integral
Line integrals extend the concept of integration from one-dimensional lines to curves in spaces, notably in the contexts of fields and differential equations. In this problem, the line integrals represent the summing up of rates of change across the boundaries of our rectangular domain.

Specifically, with \( \int_{a}^{b} \frac{\partial M}{\partial x}(x, y) \, dx \), and \( \int_{c}^{d} \frac{\partial M}{\partial y}(x, y) \, dy \), these are line integrals taken over one-dimensional paths defined by the rectangle's sides. Thus, these integrals compute the net change along an interval where one variable is kept constant.
  • Allows measurement of cumulative change across certain paths.
  • Helps demonstrate important theorems like Green's Theorem in advanced cases.
Multivariable Calculus
Multivariable calculus involves extending traditional calculus concepts, like differentiation and integration, from a single dimension to multiple dimensions. This allows us to work with functions involving several variables, providing tools for analyzing surfaces and volumes.

The exercise's application shows how the Fundamental Theorem of Calculus can be extended to functions of more than one variable. It's about understanding how the function behaves simultaneously in different directions (x and y, in this case) and how changes in these directions are captured through partial derivatives and integration.
  • Connects differential and integral calculus in higher dimensions.
  • Allows solving complex problems involving physical phenomena like fluid flow and electromagnetic fields.

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Most popular questions from this chapter

We say that two surfaces are normal at a given point if their tangent planes at that point are perpendicular to one another.Show that the pair of surfaces are normal at the given point. \(x^{2}+y^{2}+z^{2}=16\) and \(z^{2}=x^{2}+y^{2} ;(2,2,2 \sqrt{2})\)

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It seems reasonable that an increase in taxation on a commodity would decrease the production of that commodity. The following argument supports that claim. Assume that all required derivatives exist. For any \(x \geq 0\), let \(P_{0}(x)\) be the profit before taxes on \(x\) units produced. Let \(P(x, t)\) denote the profit after taxes on \(x\) units produced with \(\operatorname{tax} t\) on each unit. Assume that at any tax rate \(t\) the company will maximize its profits by producing \(f(t)\) units so that $$ \begin{aligned} &\frac{\partial P}{\partial x}(f(t), t)=0 \\ &\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)<0 \end{aligned} $$ (The conditions in (9) and (10) are just those required for the Second Derivative Test.) a. Show that \(P(x, t)=P_{0}(x)-t x\). b. Using (a), show that $$ \frac{\partial P}{\partial x}(x, t)=P_{0}^{\prime}(x)-t \quad \text { and } \quad \frac{\partial^{2} P}{\partial x^{2}}(x, t)=P_{0}^{\prime \prime}(x) $$ c. From \((9)\) and \((\mathrm{b})\), show that \(P_{0}^{\prime}(f(t))-t=0\). d. By differentiating both sides of the equation in (c) and by using (b) and (10), show that $$ f^{\prime}(t)=\frac{1}{P_{0}^{\prime \prime}(f(t))}=\frac{1}{\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)}<0 $$ (Thus the production tends to decrease as the tax rate increases.)

Find an equation of the plane tangent to the given surface at the given point. $$ \sin (x y)=2-z^{2} ;\left(\pi, \frac{1}{2},-1\right) $$

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