Problem 84 - Calculus Early Transcendentals: Pearson New International Edition

Question

Derive the formula $$ \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t=\int_{0}^{x} f(t)(x-t) d t $$ using integration by parts.

Step-by-Step Solution

Answer

Switching the order of integration and evaluating yields \( \int_{0}^{x} f(t)(x-t) \, dt \).

Step 1: Identify the Integral to Solve

Let's begin by examining the given integral: \( \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t \). We need to express this integral in a form suitable for integration by parts.

Step 2: Reverse the Order of Integration

Change the order of integration by considering the region of integration in the \(t,z\)-plane. The given integral \( \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t \) can be rewritten as \( \int_{0}^{x} \int_{z}^{x} f(z) \, dt \, dz \).

Step 3: Evaluate the Inner Integral

Evaluate the inner integral \( \int_{z}^{x} \, dt \) which simply represents the length of the interval from \(z\) to \(x\). This result is \((x-z)\). Thus, our expression becomes \( \int_{0}^{x} f(z)(x-z) \, dz \).

Step 4: Use a Dummy Variable Transformation

Switch the dummy variable from \(z\) to \(t\), as both are placeholders in the function, allowing us to rewrite the integral as \( \int_{0}^{x} f(t)(x-t) \, dt \), matching the right-hand side of the equation. The difference is superficial as it represents the same integral.

Key Concepts

Integration Techniques

Integration by parts is a fundamental method used in calculus to integrate the product of two functions. This technique is particularly useful when the integral of a product is difficult to solve directly. The method is derived from the product rule of differentiation and can be expressed using the formula:\[ \int u \, dv = uv - \int v \, du \]In this formula:

• \( u \) is a function of one of the variables, and \( dv \) is the differential of another function.
• \( v \) is the integral of \( dv \), and \( du \) is the differential of \( u \).

To apply integration by parts, you often need to decide how to choose \( u \) and \( dv \). The trick is to select \( u \) such that its derivative \( du \) is simpler than \( u \), and \( dv \) should be something you can integrate easily into \( v \).
This technique is especially valuable when dealing with integrals that arise in physics and engineering contexts, where functions are interacting with one another in complex ways.

Calculus Problems

When solving calculus problems like the one given where two nested integrals are involved, it's crucial to analyze the structure and region of integration first. The exercise involves iterated integrals which can initially appear daunting.
To tackle these:

• Visualize the integration paths and the region in the \((t, z)\)-plane to understand the bounds of integration.
• Change the order of integration if it simplifies the calculation. This involves rewriting the iterative order of two integrals to better align with easier calculation methods.

By reversing the integration order, as done in the exercise, you explore how the region is defined by constraints \(0 \leq z \leq t \leq x\), then switch to \(0 \leq z \leq x\) and \(z \leq t \leq x\). Always check back with the integral limits to ensure they correspond to the described problem.

Integral Calculus

Integral calculus focuses primarily on finding the accumulated value of a function over a range, which is critical in many scientific fields. It involves both definite integrals, like those from a limit \(a\) to \(b\), and indefinite integrals that represent a family of functions.The integral exercise demonstrates how definite integrals can be transformed by exploring algebraic expressions and geometric interpretations. Changing variables in an integral requires careful consideration, particularly when rewriting or reversing the order of integration.
Steps include:

• Identifying which variable or differential should be integrated first, and understanding the role of the function within the bounds.
• Considering "dummy variables" where functions differ only in notation or position of integration limits. This helps simplify or standardize expressions.
  

This approach reveals the underlying equivalence of expressions, showing how closely related integrals differ only by superficial variable names, not by their core computational essence.