Problem 40 - Calculus: Early Transcendentals

Question

Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves. \( y = 1.3^x \) , \( y = 2 \sqrt{x} \)

Step-by-Step Solution

Answer

Intersection points are at \( x \approx 0.89 \) and \( x \approx 4.55 \); the bounded area is about 2.92 square units.

Step 1: Graph the Functions

First, plot the two functions on the same set of axes to visualize their intersection and the region they enclose. For the first function, plot the exponential curve \( y = 1.3^x \) which starts below the y-axis and rises exponentially as \( x \) increases. For the second function, plot the curve \( y = 2\sqrt{x} \), starting from the origin (0,0) and increasing. Use graphing technology or a calculator to precisely draw these curves.

Step 2: Identify Intersection Points

Using the graph, identify the approximate points where the curves intersect. This is where their y-values are equal. Based on the graph, these intersection points are roughly around \( x = 0.89 \) and \( x = 4.55 \).

Step 3: Set Up the Integral for Area

Set up the integral to calculate the area between the curves from the identified intersection points. The formula for the area between curves is given by the integral of the upper curve minus the lower curve over the interval of intersection:\[\int_{x_1}^{x_2} (\text{upper function} - \text{lower function}) \, dx\]

Step 4: Determine Upper and Lower Functions

From the graph, determine which function is above the other. Between the intersection points, \( y = 2\sqrt{x} \) is the upper function, and \( y = 1.3^x \) is the lower function.

Step 5: Set Up and Solve the Integral

Evaluate the definite integral to find the area. The integral is set up as:\[\int_{0.89}^{4.55} (2\sqrt{x} - 1.3^x) \, dx\]This integral can be evaluated using numerical methods or a calculator. The approximate value of the area is around 2.92 square units.

Key Concepts

Graphing Functions

Graphing functions is the first crucial step in many calculus problems.
It helps visualize the behavior and shape of functions across a set of values.
In this exercise, we consider two functions: an exponential curve, \( y = 1.3^x \), and a square root function, \( y = 2 \sqrt{x} \).
To graph these functions, follow these steps:

• For \( y = 1.3^x \), note that it is an exponential growth function starting below the y-axis, and it rises sharply as \( x \) increases.
• For \( y = 2 \sqrt{x} \), it begins at the origin and increases, but more gradually compared to the exponential curve.
• Use graphing tools or calculations to plot these curves on the same axes.

The graph helps visualize where the two functions may intersect and how they enclose any area.

Intersection of Curves

Finding the intersection of curves is about identifying where two functions have the same output value for different input values.
This corresponds to the \( x \)-coordinates where the graphs intersect.To find the intersection:

• Graphically locate where \( y = 1.3^x \) and \( y = 2 \sqrt{x} \) intersect.
• Through graphing or using a calculator's intersection feature, estimate these points.
• In this case, the intersection points are approximately \( x = 0.89 \) and \( x = 4.55 \).

This provides the x-values necessary for setting up integrals to find enclosed areas.

Area Between Curves

The area between curves represents the region enclosed by the two functions.
Understanding this concept involves calculating the vertical space between the curves within specified limits. The area is typically found using calculus techniques.For this exercise:

• The region between \( y = 1.3^x \) and \( y = 2 \sqrt{x} \) is bounded by the intersection points \( x = 0.89 \) and \( x = 4.55 \).
• To find the area, one must set up an integral that accounts for the height difference between these functions over the range.

This step precedes setting up a definite integral to compute the specific area.

Integral Setup

Setting up an integral to find the area between curves is a fundamental calculus skill.
The integral is configured to measure the typical "slice" of area between two curves over an interval.Here’s how to set it up:

• Identify which function is on top, which in this case is \( y = 2 \sqrt{x} \).
• Recognize the lower function, here \( y = 1.3^x \).
• Set up the integral as \( \int_{0.89}^{4.55} (2\sqrt{x} - 1.3^x) \, dx \).

The integral represents a sum of differential vertical strips between these two functions from one intersection to the other.

Numerical Integration

Numerical integration techniques are used when calculating exact integrals is difficult or approximations are sufficient.
These techniques allow approximation of the value of definite integrals where analytical solutions are complex or cumbersome.For this problem:

• Given the integral \( \int_{0.89}^{4.55} (2\sqrt{x} - 1.3^x) \, dx \), exact evaluation might be tedious.
• Numerical methods, like the trapezoidal rule or Simpson’s rule, simplify this process.
• Using calculators, we approximate the area to be about 2.92 square units.

Such methods are invaluable in practical applications where quick, reliable estimates are necessary.