Problem 52 - Calculus Volume 3

Question

Show that $$ \int_{a}^{b} \int_{c}^{d} y f(x)+x g(y) d y d x=\frac{1}{2}\left(d^{2}-c^{2}\right)\left(\int_{a}^{b} f(x) d x\right)+\frac{1}{2}\left(b^{2}-a^{2}\right)\left(\int_{c}^{d} g(y) d y\right) $$

Step-by-Step Solution

Answer

The double integral evaluates to the sum of two separate integrals: \( \frac{1}{2}(d^2 - c^2) \int_{a}^{b} f(x) dx + \frac{1}{2}(b^2 - a^2) \int_{c}^{d} g(y) dy \).

Step 1: Expand the Inner Integral

First, consider the inner integral \( \int_{c}^{d} y f(x) + x g(y) \ dy \). We can separate this integral into two separate integrals: \( \int_{c}^{d} y f(x) \ dy + \int_{c}^{d} x g(y) \ dy \). This simplification will help us tackle each part individually.

Step 2: Solve \( \int_{c}^{d} y f(x) \ dy \)

The integral \( \int_{c}^{d} y f(x) \ dy \) is a function of \( x \) because \( f(x) \) is considered a constant when integrating with respect to \( y \). The integration yields \( f(x) \cdot \left[ \frac{1}{2}y^2 \right]_{c}^{d} = f(x) \cdot \frac{1}{2}(d^2 - c^2) \).

Step 3: Solve \( \int_{c}^{d} x g(y) \ dy \)

Here, \( \int_{c}^{d} x g(y) \ dy = x \int_{c}^{d} g(y) \ dy \). This is because \( x \) is treated as a constant during integration with respect to \( y \). Thus, the result is \( x \cdot G \), where \( G = \int_{c}^{d} g(y) \ dy \).

Step 4: Evaluate the Outer Integral for Each Part

Now, we focus on the outer integrals. For the first part, integrate \( \frac{1}{2}(d^2 - c^2) f(x) \) with respect to \( x \) from \( a \) to \( b \):\[ \int_{a}^{b} \frac{1}{2}(d^2 - c^2)f(x) \ dx = \frac{1}{2}(d^2 - c^2) \cdot \int_{a}^{b} f(x) \ dx \].For the second outer integral, integrate \( x G = x \int_{c}^{d} g(y) \ dy \) with respect to \( x \) from \( a \) to \( b \):\[ G \int_{a}^{b} x \ dx = G \cdot \left[ \frac{1}{2}x^2 \right]_{a}^{b} = \frac{1}{2}G(b^2 - a^2) \].

Step 5: Combine Results

Combine the results of the outer integrals to form the entire double integral. Upon substitution, the expression becomes:\[ \frac{1}{2}(d^2 - c^2) \int_{a}^{b} f(x) \ dx + \frac{1}{2}(b^2 - a^2) \int_{c}^{d} g(y) \ dy \].This matches the given expression in the problem statement, confirming the equation is correct.

Key Concepts

Multiple Integration

Multiple integration is an extension of single-variable integration to functions of multiple variables. It allows us to compute the accumulated quantity over a multidimensional region. In the case of our exercise, we deal with a double integral, which involves two layers of integration.

Consider we have a function of two variables, such as \( f(x, y) \). We can use multiple integration to integrate this function over a specific rectangular region defined by \([a, b] \) for \(x\) and \([c, d] \) for \(y\). This two-step integration can be visualized as:

• Integrating with respect to \(y\), treating \(x\) as a constant (\(\int_c^d f(x, y) \ dy\)).
• Subsequently, integrating the result with respect to \(x\) (\(\int_a^b (\text{result from previous step}) \ dx\)).

By breaking down the integration process in this ordered manner, we can evaluate complex functions over multiple dimensions easily, making it a valuable tool in various fields like physics and engineering.

Integral Calculus

Integral calculus forms one of the foundational pillars of calculus. It's concerned with the concept of integration, which is essentially finding the antiderivative or the accumulation of quantities.

In the realm of double integrals, integral calculus expands to cover functions of two or more variables. We employ it to determine values that describe areas and volumes under curves or surfaces. It not only helps to compute these quantities but also unveils properties and relations between functions.

In our problem, integral calculus allows us to break down a complicated integrand like \( yf(x) + xg(y) \) into simpler parts. By separating and acknowledging the rules of integration, such as treating one variable constant while integrating with respect to another, we can solve each part effectively. This

• utilizes techniques like substitution and partial fraction integration,
• applies fundamental theorems that connect derivatives with integrals to ease the integration process.

Furthermore, knowing the behavior of polynomials and other functions under integration simplifies working through such exercises.

Mathematical Proofs

Mathematical proofs are methods used to demonstrate the truth of an assertion or statement. Proving mathematical equations like the one in the exercise involves validating each transformation and logical step to ensure the conclusion reliably follows from the assumptions.

In our exercise, proving the double integral expression involves several key proof strategies:

• Decomposition: Break down the complex expression into manageable parts.
• Substitution: Use algebraic manipulation to transform terms into more mathematically convenient formats.
• Reconstruction: Combining results from individual integrations to rebuild and verify the original equation.

Proofs not only confirm the validity of mathematical equations but also deepen our conceptual understanding of the relationships between mathematical operations. These logical structures act as the bedrock of mathematical theory, ensuring that solutions are not only correct but comprehensible and reliable. Engaging with proofs fosters deeper analytical thinking and problem-solving skills. Reacting this way to mathematical challenges builds a robust understanding required for complex problem solving.