Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1–10, evaluate the iterated integrals. $$ \int_{-3}^{7} \int_{0}^{2 x} \int_{y}^{x-1} d z d y d x $$

Step-by-Step Solution

Answer

The evaluated iterated integral is -40.

Step 1: Evaluate the innermost integral with respect to z

The innermost integral is \( \int_{y}^{x-1} d z \).To solve this, we find the antiderivative of 1 with respect to \(z\), which is \(z\). With limits of integration from \(y\) to \(x-1\), the result is:\[(z \big|_{y}^{x-1}) = (x-1) - y\]

Step 2: Evaluate the middle integral with respect to y

Now, we need to integrate the result from Step 1 with respect to \(y\):\[ \int_{0}^{2x} ((x-1) - y) \, d y\]First, separate the terms:\[\int_{0}^{2x} (x-1) \, d y - \int_{0}^{2x} y \, d y\]For the first integral:\[(x-1)y \big|_{0}^{2x} = (x-1)(2x) - 0 = 2x(x-1)\]For the second integral:\[\frac{1}{2}y^2 \big|_{0}^{2x} = \frac{1}{2}(2x)^2 - 0 = 2x^2\]Subtract these results:\[2x(x-1) - 2x^2 = 2x^2 - 2x - 2x^2 = -2x\]

Step 3: Evaluate the outermost integral with respect to x

The remaining integral is:\[\int_{-3}^{7} -2x \, d x\]The antiderivative of \(-2x\) is \(-x^2\). Evaluate it over the interval \([-3, 7]\):\[(-x^2) \big|_{-3}^{7} = (-(7)^2) - (-(3)^2) = -49 - (-9) = -49 + 9 = -40\]

Step 4: Final Step: Compile the solution

Through our calculations, we have determined that the value of the iterated integral is \(-40\).

Key Concepts

Understanding Multiple Integrals

Multiple integrals extend the concept of integration to functions with more than one variable. Instead of just integrating with respect to one variable, we consider

• two variables (double integrals),
• three variables (triple integrals),
• and so on for higher dimensions.

This is crucial in calculus, especially for calculating areas, volumes, and centers of mass in multi-dimensional spaces. When evaluating a multiple integral, such as our given iterated integral, we proceed by integrating one variable at a time.
The idea is to simplify the problem step-by-step, integrating from the innermost to the outermost variable. This process makes complex multi-dimensional calculus problems more manageable by breaking them down into a series of simpler, single-variable integrals.

Significance of Antiderivative in Integration

The antiderivative, also known as the indefinite integral, forms the backbone of any integration process. When we perform integration, we essentially reverse the process of differentiation.
In simpler terms, finding the antiderivative is like going backwards from a derivative to the original function.

• For a function with respect to one variable, integrating will give us its antiderivative.
• In our iterated integral, for instance, we first find the antiderivative of 1 with respect to z, yielding z.

The antiderivative plays a key role in determining areas under curves or between curves. Without it, the fundamental purpose of integration would falter, as it provides us with a way to perform the summation of infinitely small quantities over a given range of a function.

Setting and Understanding Integration Limits

Integration limits define the range over which we integrate a function. The boundaries of these limits are crucial for solving any integral, and they require careful setup:

• Lower and Upper Limits: In our exercise, we have -3 to 7 for x, 0 to 2x for y, and y to x-1 for z.
• Iterated Integrals: Require evaluation from the innermost to the outermost layer, respecting each variable's limit.

Limits dictate what portion of the function we sum over and are responsible for the structure of the integral setup over specific intervals. Correctly interpreting and applying these limits ensures the result accurately represents the problem being solved, such as calculating part of a volume or area in physics or engineering challenges.

Tackling Calculus Problems with Iterated Integrals

Handling calculus problems involving iterated integrals can seem daunting at first, but systematic approaches ease the process:

• Break Down the Problem: Always begin by resolving the innermost integral and gradually move outward.
• Step-by-Step Calculation: Avoid rushing. Solve each integral separately and ensure individual steps like subtraction are correctly managed.
• Keep Track of Limits and Variables: Mismanaging limits or mixing up variables is a common error to watch for.

By treating these complex problems in layers, we apply basic integral concepts repeatedly. This problem-solving approach is especially handy when variables and their limits are interdependent.
Ultimately, tackling iterated integrals strengthens one’s mathematical maturity, enabling comprehension of more advanced calculus concepts.