Problem 57 - Calculus with Concepts in Calculus

Question

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 \(a \leq x \leq b\)

Step-by-Step Solution

Answer

The integral property holds by the Fundamental Theorem of Calculus.

Step 1: 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.

Step 2: 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, \( \int_{a}^{b} \frac{\partial M}{\partial x}(x, y) \, dx = M(b, y) - M(a, y) \).

Step 3: 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.

Step 4: 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.

Step 5: 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.

Key Concepts

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.