Problem 40 - Thomas Calculus

Question

If \(f(x, y)\) is continuous over \(R : a \leq x \leq b, c \leq y \leq d\) and $$F(x, y)=\int_{a}^{x} \int_{c}^{y} f(u, v) d v d u$$ on the interior of \(R,\) find the second partial derivatives \(F_{x y}\) and \(F_{y x}\)

Step-by-Step Solution

Answer

Both \(F_{xy}(x, y)\) and \(F_{yx}(x, y)\) are equal to \(f(x, y)\).

Step 1: Define the Region of Integration

Given that the function \(f(x, y)\) is continuous over the rectangle \(R : a \leq x \leq b, c \leq y \leq d\), the double integral associated with \(F(x, y)\) is computing the area over this specified region using \(f(u, v)\).

Step 2: Understand the Function \(F(x, y)\)

The function \(F(x, y)\) is defined as:\[ F(x, y) = \int_{a}^{x} \int_{c}^{y} f(u, v) \, dv \, du \]This represents the cumulative effect or aggregation of \(f(u, v)\) over the region from \((a, c)\) to \((x, y)\).

Step 3: Find the Partial Derivative \(F_x(x, y)\)

To find \(F_{xy}(x, y)\), first differentiate \(F(x, y)\) with respect to \(x\):\[ F_x(x, y) = \frac{d}{dx} \left( \int_{a}^{x} \int_{c}^{y} f(u, v) \, dv \, du \right) = \int_{c}^{y} f(x, v) \, dv \]Here, the differentiation under the integral with respect to \(x\) replaces \(u\) with \(x\).

Step 4: Differentiate \(F_x(x, y)\) with Respect to \(y\)

Continue to find \(F_{xy}(x, y)\) by differentiating \(F_x(x, y)\) with respect to \(y\):\[ F_{xy}(x, y) = \frac{d}{dy} \left( \int_{c}^{y} f(x, v) \, dv \right) = f(x, y) \]This is because differentiation with respect to \(y\) simply evaluates the inner integral with respect to the new variable at \(v = y\).

Step 5: Find the Partial Derivative \(F_y(x, y)\)

Next, differentiate \(F(x, y)\) directly with respect to \(y\):\[ F_y(x, y) = \frac{d}{dy} \left( \int_{a}^{x} \int_{c}^{y} f(u, v) \, dv \, du \right) = \int_{a}^{x} f(u, y) \, du \]This result comes by differentiating under the integral, assuming \(v\) goes from \(c\) to \(y\).

Step 6: Differentiate \(F_y(x, y)\) with Respect to \(x\)

Finally, find \(F_{yx}(x, y)\) by differentiating \(F_y(x, y)\) with respect to \(x\):\[ F_{yx}(x, y) = \frac{d}{dx} \left( \int_{a}^{x} f(u, y) \, du \right) = f(x, y) \]This follows via direct differentiation of the integral evaluated at \(x\).

Step 7: Conclusion: Equate the Mixed Partial Derivatives

Both mixed partial derivatives \(F_{xy}(x, y)\) and \(F_{yx}(x, y)\) have been shown to equal \(f(x, y)\), confirming that the function is smooth and continuous over the region.

Key Concepts

Continuity of Functions

The concept of continuity is a fundamental characteristic in calculus. A function is said to be continuous at a point if, intuitively, you can draw its graph without lifting your pen from the paper. In mathematical terms, a function \( f(x, y) \) is continuous over a region \( R \) if, for every point \((x_0, y_0)\) in that region, the function value \( f(x_0, y_0) \) approaches \( f(x, y) \) as \( (x, y) \) approaches \( (x_0, y_0) \). This means there are no breaks, jumps, or holes in the graph of the function within that region.

Continuity is a crucial requirement when working with mixed partial derivatives. If a function is continuous, we can interchange the order of differentiation, leading to equal mixed partial derivatives, like \( F_{xy} = F_{yx} \). Thus, continuity assures us that the order in which we differentiate doesn't affect the result, making calculations neat and predictable. This property is essential when evaluating functions over complex regions using integrals.

Double Integrals

Double integrals are a way to calculate the volume under a surface over a specified region. For a function \( f(x, y) \), the double integral over region \( R \) can be expressed as \( \int \int_R f(x, y) \; dx \; dy \). When we compute double integrals, we essentially find the accumulation of the function values over the entire region, giving us a sense of the total effect.

In the context of the problem, double integrals connect to how we compute \( F(x, y) \), the function defined by integrating first with respect to \( v \) and then with respect to \( u \). This multistep integration reflects how we handle multidimensional functions. The continuity of \( f(x, y) \) ensures the double integral is well-defined, allowing us to smoothly evaluate the function's accumulation over the region.

Differentiation under the Integral Sign

Differentiation under the integral sign is an advanced technique that allows us to differentiate an integral with variable limits. This might sound complex, but it gives us powerful tools to solve certain problems more efficiently. The key idea is that, under certain conditions, we can pass differentiation inside the integral. This is rooted in Leibniz's rule.

In our exercise, when we found \( F_x(x, y) \) or \( F_y(x, y) \), we used this technique. By differentiating \( F(x, y) \) with respect to one variable while the other still bounds the integral, we transform an accumulation problem into a derivative one. For example, differentiating with respect to \( x \) when \( y \) bounds the inner integral simplifies the computation by holding \( y \) as a constant boundary, reducing the problem to evaluating the function at given points.

Fundamental Theorem of Calculus for Line Integrals

The Fundamental Theorem of Calculus for line integrals extends the traditional theorem from one-dimensional calculus to line integrals over curves. It states that if a vector field is the gradient of some scalar function, the line integral of the vector field over some curve depends only on the values of this scalar function at the endpoints of that curve.

In the context of our original exercise, while not directly used, this concept underlies the logic of evaluating \(F(x, y)\) and finding mixed partial derivatives. Each derivative can be viewed as measuring the change in the accumulated area defined by \( F(x, y) \) as we move along the axes. Understanding this broader perspective helps us realize that the evaluations of these integrals respond consistently to the changes within their regions, similar to how integrating over paths uses endpoints to simplify calculations.