Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. \(\int_{0}^{1} \int_{0}^{2} \frac{y}{1+x^{2}} d y d x\)

Step-by-Step Solution

Answer

The value of the iterated integral is \( \frac{\pi}{2} \).

Step 1: Setup the Iterated Integral

The given iterated integral is \( \int_{0}^{1} \left( \int_{0}^{2} \frac{y}{1+x^{2}} \, d y \right) \, d x \). We need to solve the inner integral first, which involves the variable \( y \).

Step 2: Evaluate the Inner Integral

Focus on the inner integral \( \int_{0}^{2} \frac{y}{1+x^{2}} \, d y \). Treat \( \frac{1}{1+x^{2}} \) as a constant with respect to \( y \), so the integral becomes \( \frac{1}{1+x^{2}} \int_{0}^{2} y \, d y \).

Step 3: Integrate with Respect to y

Evaluate \( \int_{0}^{2} y \, d y \), which is a basic integral of the form \( \int y^n \, dy = \frac{y^{n+1}}{n+1} \). Thus, \( \int_{0}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = 2 \).

Step 4: Multiply by Constant Factor

Multiply the result of the inner integral by the constant factor \( \frac{1}{1+x^{2}} \), giving \( \frac{2}{1+x^{2}} \). This is now the result of the inner integral substitution back in the outer integral.

Step 5: Evaluate the Outer Integral

Solve the outer integral \( \int_{0}^{1} \frac{2}{1+x^{2}} \, d x \). This is equivalent to \( 2 \int_{0}^{1} \frac{1}{1+x^{2}} \, d x \), which is a standard integral that results in \( \arctan(x) \).

Step 6: Integrate with Respect to x

Evaluate \( 2 \int_{0}^{1} \frac{1}{1+x^{2}} \, d x = 2 \left[ \arctan(x) \right]_{0}^{1} = 2 \left( \arctan(1) - \arctan(0) \right) \).

Step 7: Calculate the Arctangent Values

Calculate the values: \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(0) = 0 \). Thus, the evaluation becomes \( 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2} \).

Key Concepts

Multiple Integration

In calculus, multiple integration refers to the process of integrating a function with more than one variable, like our given example \[ \int_{0}^{1} \int_{0}^{2} \frac{y}{1+x^{2}} \, d y \, d x \]This expression involves two integrals performed sequentially. It essentially sums up infinitely small products over a defined region, enabling calculation of volumes under surfaces.

• The inner integral is solved first, treating remaining terms as constants to extract a result at each point.
• The outer integral then integrates this result over another dimension, completing the process.

Multiple integration is foundational in multivariable calculus, supporting applications like physics and engineering to model real-world phenomena. Breaking it into understanding individual variables simplifies the complexity.

Calculus

Calculus is the mathematical study of continuous change, comprising differentiation and integration, crucial for understanding advanced mathematics. In calculus, integration is used to calculate areas under curves, volumes, and other scenarios involving summation of infinitesimals. Integration is often split into single, double (as in our exercise), and even triple integrals, depending on the number of variables involved. Each type further deepens our understanding of how variables interact, building from single-variable functions to complex systems. The provided exercise is a double integral, showcasing how calculus extends to consider both dimensions for the problem-solving process. Generally,

• Single integrals focus on area under a curve.
• Double integrals evaluate volumes under surfaces.
• Triple integrals consider hyper-volumes in higher-dimensional spaces.

Mastering these concepts leads to powerful tools that interpret multidimensional spaces practically.

Integration Techniques

Integration techniques simplify the process of solving complex integrals, such as iterated integrals. Several methods are used, with choice often dictated by the function type and form.In our example, separation of variables is applied by treating terms as constants when solving the inner integral:

• First, \(\int_{0}^{2} \frac{y}{1+x^{2}} \, d y\) formed by separating \(\frac{1}{1+x^{2}}\) is constant with respect to \(y\).
• Then, simple integration rules, like the power rule, were used to evaluate \(\int y \, dy\).

Finally, recognizing the standard form in the outer integral \(\int_{0}^{1} \frac{1}{1+x^{2}} \, d x \), leads to solving via a direct antiderivative \(\arctan(x)\), showcasing the efficacy of recognizing integral patterns.Understanding and mastering different techniques can ease work with complex problems and ensure each step aligns with mathematical principles, reducing errors.