Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. $$ \int_{1}^{5} \int_{0}^{x} \frac{3}{x^{2}+y^{2}} d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is \( \frac{3\pi}{4} \ln 5 \).

Step 1: Understand the Integral

We are asked to evaluate the double integral \( \int_{1}^{5} \int_{0}^{x} \frac{3}{x^{2}+y^{2}} d y d x \). The order of integration is \( dy \) then \( dx \). This means that for each fixed \( x \), we first integrate with respect to \( y \), where \( y \) ranges from 0 to \( x \).

Step 2: Integrate with respect to y

We focus on the inner integral, \( \int_{0}^{x} \frac{3}{x^{2}+y^{2}} d y \). Use the substitution \( u = y \), so that \( du = dy \). The integral becomes \( 3 \int_{0}^{x} \frac{1}{x^{2} + u^{2}} du \), which can be recognized as an arctangent integral.

Step 3: Solve the Inner Integral

The integral \( \int \frac{1}{x^2 + u^2} \, du \) is \( \frac{1}{x} \tan^{-1} \left( \frac{u}{x} \right) + C \). Applying this to our bounds from 0 to \( x \), we get:\[3 \cdot \left( \frac{1}{x} \tan^{-1} \left( \frac{x}{x} \right) - \frac{1}{x} \tan^{-1} \left( 0 \right) \right)\]This simplifies to \( 3 \cdot \left( \frac{1}{x} \cdot \frac{\pi}{4} \right) = \frac{3\pi}{4x} \).

Step 4: Integrate with respect to x

Now integrate \( \int_{1}^{5} \frac{3\pi}{4x} \, dx \). This integral becomes \( \frac{3\pi}{4} \int_{1}^{5} \frac{1}{x} \, dx \), which is a logarithmic integral.

Step 5: Solve the Outer Integral

The integral \( \int \frac{1}{x} \, dx \) is \( \ln |x| + C \). Evaluating from 1 to 5 gives:\[\frac{3\pi}{4} \cdot (\ln 5 - \ln 1)\]Since \( \ln 1 = 0 \), this simplifies to \( \frac{3\pi}{4} \cdot \ln 5 \).

Step 6: Write the Final Answer

The value of the iterated integral is \( \frac{3\pi}{4} \ln 5 \).

Key Concepts

Double Integrals

Double integrals are an extension of single-variable integrals, allowing us to compute the volume under a surface over a certain region. In this exercise, we have a double integral, \( \int_{1}^{5} \int_{0}^{x} \frac{3}{x^2+y^2} \, dy \, dx \). Here, the order of integration is crucial. We first integrate with respect to \( y \) and then with respect to \( x \). This means that for every \( x \) value, we are calculating the area from \( y = 0 \) to \( y = x \).

• Inner Integral: Represents the integral over \( y \) for a fixed \( x \).
• Outer Integral: It sums up the values along the \( x \) axis, over the interval from \( 1 \) to \( 5 \).

Mastering double integrals is essential in multivariable calculus, as they model various real-world phenomena ranging from physical to financial situations.

Integration Techniques

When solving iterated integrals, choosing the right integration techniques is vital for success. Here, we handle two types of integrals: an arctangent integral and a logarithmic integral. In general, identifying the type of function involved in the integral can guide you:

• Substitution Method: Useful when simplifying the function or its limits via a substitution variable, like in this problem where \( u = y \).
• Recognizing Standard Forms: This is crucial when solving the inner integral related to arctangents.

Resorting to standard integral forms or substitutions saves time and minimizes errors. Practicing these techniques solidifies your calculus skills.

Arctangent Integral

Arctangent integrals arise when integrating expressions such as \( \frac{1}{x^2 + y^2} \). These integrals result in a form involving the inverse tangent function, \( \tan^{-1} \).
When we encounter \( \int \frac{1}{x^2 + u^2} \, du \), it translates to \( \frac{1}{x} \tan^{-1} \left( \frac{u}{x} \right) + C \).

• In our problem, the bounds \(0\) to \( x \) were applied, simplifying it to \( \frac{3\pi}{4x} \).
• Understanding Inverse Functions: Arctangent is crucial here, typically returning angles whose tangents match the given ratios.

Recognizing these forms can streamline the solution process and helps in dealing with trigonometric integrals effectively.

Logarithmic Integral

Logarithmic integrals are solved using natural logarithms, benefiting us when integrating functions of the form \( \int \frac{1}{x} \, dx \). This results in \( \ln |x| + C \). When solving our given integral \( \int_{1}^{5} \frac{1}{x} \, dx \), we calculate \( \ln 5 - \ln 1 \).
Since \( \ln 1 = 0 \), it simplifies to \( \ln 5 \).

• Natural Logarithms: Serve as the fundamental tool in handling exponential and growth-related functions.
• Evaluating Definite Integrals: Critical in transitioning from antiderivatives to actual numerical results.

Logarithmic connections appear often in calculus, providing solutions in various models, especially those involving decay and growth processes.