Problem 24 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the region of integration

The limits given are \( x \) from 1 to 4 and \( y \) from 0 to \( \sqrt{x} \). This means the region of integration is bounded by \( x = 1 \), \( x = 4 \), \( y = 0 \), and \( y = \sqrt{x} \).

Step 2: Sketch the region in the xy-plane

The region can be sketched on an xy-plane. The line \( y = \sqrt{x} \) starts from (1,1) and extends to a higher value as \( x \) reaches 4 (at the point \( x=4, y=2\)). The area is under the curve \( y = \sqrt{x} \) and above the line \( y = 0 \), between \( x = 1 \) and \( x = 4 \).

Step 3: Evaluate the inner integral

The inner integral is \( \int_{0}^{\sqrt{x}} \frac{3}{2} e^{y / \sqrt{x}} \, d y \). Using substitution \( u = y/\sqrt{x} \), \( du = \frac{1}{\sqrt{x}} dy \), transform the limits of integration to \( 0 \) and \( 1 \). Integrate: \( \int_{0}^{1} \frac{3}{2} e^{u} \sqrt{x} \, du = \frac{3}{2} \left[\sqrt{x} e^{u} \right]_{0}^{1} \). Evaluate the bounds to get \( \frac{3}{2} (\sqrt{x} e^{1} - \sqrt{x} e^{0}) = \frac{3}{2} \sqrt{x} (e - 1) \).

Step 4: Evaluate the outer integral

The outer integral is \( \int_{1}^{4} \frac{3}{2} \sqrt{x} (e - 1) \, dx \). Factor out constants: \( \frac{3}{2}(e - 1) \int_{1}^{4} \sqrt{x} \, dx \). Change \( \sqrt{x} \) to \( x^{1/2} \) and integrate: \( \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \). Evaluate it from 1 to 4: \( \frac{2}{3}[(4^{3/2}) - (1^{3/2})] = \frac{2}{3}[8 - 1] = \frac{14}{3} \).

Step 5: Compute the final result

Multiply the result by the constant factor: \( \frac{3}{2}(e - 1) \times \frac{14}{3} = 7(e - 1) \).

Key Concepts

Region of Integration

In double integration, understanding the region of integration is key to setting up the integral correctly. In our exercise, we are given the bounds for the variables \(x\) and \(y\):

• \(x\) ranges from 1 to 4
• \(y\) ranges from 0 to \(\sqrt{x}\)

This specifies a certain area on the Cartesian plane. The line \(y = \sqrt{x}\) is a curve that begins at the point (1,1) and "grows" upwards reaching (4,2) as x increases. The region of integration is formed by this curve above the x-axis and the vertical lines \(x = 1\) and \(x = 4\). Sketching the region helps to visually identify this area as it is crucial to understanding where the integration will apply.

Change of Variables

When tackling the inner integral, changing variables is often a strategic move to simplify computation. In the given problem, we use substitution to evaluate the inner integral:

• Let \(u = \frac{y}{\sqrt{x}}\), which implies that as \(y\) moves from 0 to \(\sqrt{x}\), \(u\) moves from 0 to 1.
• This substitution also alters the differential: \(du = \frac{1}{\sqrt{x}} \, dy\).

Implementing this change of variables effectively transforms the integral: \(\int_{0}^{\sqrt{x}} \frac{3}{2} e^{y/\sqrt{x}} \, dy\) becomes \(\int_{0}^{1} \frac{3}{2} \sqrt{x} e^{u} \, du\). Such substitutions are instrumental in simplifying complicated integrals into more manageable forms.

Integration Limits

Establishing correct integration limits is essential for accurate computation. In our case, the integration is done in two steps, often referred to as iterated integration:

• The inner integration for \(dy\) requires limits from \(0\) to \(\sqrt{x}\).
• The outer integration for \(dx\) ranges from \(1\) to \(4\).

These limits ensure that the integral precisely covers the intended area under the curve \(y = \sqrt{x}\). When we compute the outer integral, we integrate over \(x\), treating the outcome of the inner integral as a constant. A meticulous approach to defining these limits is crucial to accurately solving integrals and obtaining the right results, as missteps here can lead to significant errors.