Problem 24 - Thomas Calculus

Question

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{1}^{4} \int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} d y d x \end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( 7(e-1) \).

Step 1: Understand the Limits of Integration

The given integral is a double integral over the region defined by the limits: \( x \) goes from 1 to 4, and for each \( x \), \( y \) goes from 0 to \( \sqrt{x} \). This means that the region of integration is bounded below by \( y = 0 \) and above by the curve \( y = \sqrt{x} \) on the domain \( 1 \leq x \leq 4 \).

Step 2: Sketch the Region of Integration

Plot the region in the xy-plane: The line \( x = 1 \) and \( x = 4 \) forms the vertical boundaries, and \( y = \sqrt{x} \) forms a curve starting from \( (1, 1) \) to \( (4, 2) \), above the line \( y = 0 \). The area of interest is the shaded region below the curve \( y = \sqrt{x} \) and between \( x = 1 \) and \( x = 4 \).

Step 3: Evaluate the Inner Integral

Performing the integration with respect to \( y \), treat \( x \) as a constant: \[ \int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} \, dy \].To integrate, substitute \( u = \frac{y}{\sqrt{x}} \) so that \( du = \frac{1}{\sqrt{x}} \, dy \), or equivalently, \( dy = \sqrt{x} \, du \). The limits of \( u \) change from 0 to 1. Therefore, the integral becomes:\[ \frac{3\sqrt{x}}{2} \int_{0}^{1} e^{u} \, du.\]Evaluating this integral results in:\[ \frac{3\sqrt{x}}{2} [e^{u}]_{0}^{1} = \frac{3\sqrt{x}}{2} (e - 1).\]

Step 4: Evaluate the Outer Integral

Now integrate with respect to \( x \):\[ \int_{1}^{4} \frac{3\sqrt{x}}{2} (e - 1) \, dx.\]Since \( \frac{3(e-1)}{2} \) is a constant, factor it out:\[ \frac{3(e-1)}{2} \int_{1}^{4} \sqrt{x} \, dx.\]The integral of \( \sqrt{x} \) is \( \frac{2}{3}x^{3/2} \). Evaluate this from 1 to 4:\[ \frac{3(e-1)}{2} \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4} = \frac{3(e-1)}{2} \left( \frac{2}{3} \left[(4)^{3/2} - (1)^{3/2}\right] \right).\]Calculate: \( 4^{3/2} = 8 \) and \( 1^{3/2} = 1 \), results in:\[ \frac{2}{3} (8 - 1) = \frac{14}{3}.\]Therefore, the value of the outer integral is:\[ \frac{3(e-1)}{2} \cdot \frac{14}{3} = 7(e-1).\]

Step 5: Final Calculation and Result

Multiply through by the constant factors to get the final result:\[ 7(e-1).\] Thus, the evaluated integral is \( 7(e-1) \).

Key Concepts

Region of Integration

In the context of a double integral, the **region of integration** refers to the area in the coordinate plane over which the integration occurs. Understanding the region of integration is crucial for visualizing the problem and ensuring accurate computation of the integral.
In our specific exercise, the region is bounded on the x-axis by the lines \( x = 1 \) and \( x = 4 \). On the y-axis, it is bounded below by \( y = 0 \), and above by the curve \( y = \sqrt{x} \).
This setup often involves sketching or at least imagining the area involved, so you understand the boundaries well.

• The limits for \( y \) depend on the value of \( x \). Unlike a regular bounded rectangle or square, this introduces a dynamic where \( y \) varies with \( \sqrt{x} \).
• Visualizing this region as a part of the xy-plane helps see that it forms a triangle-like shape, curving upwards between \( y = 0 \) and \( y = \sqrt{x} \), all while \( x \) ranges from 1 to 4.

This visualization is the foundation for setting up our double integral correctly.

Limits of Integration

**Limits of integration** delineate the boundaries for the integration process. These limits tell us where to start and stop integrating along each axis.
For a double integral, the limits are specified for both variables, such as in our integral \( \int_{1}^{4} \int_{0}^{\sqrt{x}} f(x, y) \, dy \, dx \).

• The outer limits \( 1 \leq x \leq 4 \) mean we integrate in the x-direction from 1 to 4. It defines a horizontal span along the x-axis.
• For each fixed \( x \), the inner limits \( 0 \leq y \leq \sqrt{x} \) tell us how far to integrate in the y-direction, which varies depending on \( x \).

This combination of inner and outer limits requires careful interpretation, as the limits of one integral (y) depend on the other variable (x). This allows for regions defined by curves, not just straight lines.

Integration by Substitution

**Integration by substitution** is a technique similar to the substitution used in regular calculus integration. It involves changing variables to simplify the integration process.
During this exercise, when computing the inner integral, the substitution \( u = \frac{y}{\sqrt{x}} \) is used to replace \( y \) with \( u \), making the integral easier to handle.

• We compute \( du = \frac{1}{\sqrt{x}} \, dy \), allowing us to express \( dy = \sqrt{x} \, du \).
• The limits change from 0 to \( \sqrt{x} \) to 0 to 1 after substitution, simplifying the problem.

Substitution transforms the integrand \( e^{y / \sqrt{x}} \) into \( e^u \), a straightforward function to integrate. The main goal here is to simplify the integral to a form that is easier to compute.

Calculus

Calculus encompasses the theory and practice of integration, differentiation, and their applications. It's the mathematical backbone of continuous change and accumulation.
Double integrals, a key topic in multivariable calculus, extend these ideas over regions in two dimensions.
This exercise demonstrates the process of evaluating double integrals involving variables of both x and y. Each variable can represent a different dimension, intensifying the need for accurate technique.

• Integration: It's the process of finding the accumulation of quantities, like area under a curve.
• Substitution: A basic calculus technique that simplifies difficult integrals.

These concepts work together to solve complex problems involving areas defined not just by straight lines but by curves and multi-dimensional shapes. Calculus captures changes not only in one dimension but across surfaces, as shown by our example problem.