Problem 5 - Thomas Calculus

Question

In Exercises 1–8, sketch the region bounded by the given lines and curves. Then express the region’s area as an iterated double integral and evaluate the integral. The curve \(y=e^{x}\) and the lines \(y=0, x=0,\) and \(x=\ln 2\)

Step-by-Step Solution

Answer

The area of the region is 1.

Step 1: Sketch the Region

First, plot the curve and lines on the graph. The curve given is \( y = e^x \), which is an exponential curve. The lines are \( y = 0 \) (the x-axis), \( x = 0 \) (the y-axis), and \( x = \ln 2 \). The region is the area on the graph that is enclosed by these lines and the curve.

Step 2: Determine the Limits of Integration

Given the boundaries, the region of interest is between \( x = 0 \) and \( x = \ln 2 \). The upper boundary of the region is defined by the curve \( y = e^x \), and the lower boundary is the line \( y = 0 \). Therefore, the limits for \( y \) are \( 0 \) to \( e^x \) for each value of \( x \).

Step 3: Set Up the Double Integral

The area can be calculated using a double integral. The double integral is set up as follows: \[ \int_{x = 0}^{\ln 2} \int_{y = 0}^{e^x} \, dy \, dx \].

Step 4: Calculate the Inner Integral

Compute the inner integral with respect to \( y \):\[ \int_{y = 0}^{e^x} \, dy = e^x \]This results because the integral of \( dy \) is simply \( y \), evaluated from \( 0 \) to \( e^x \).

Step 5: Evaluate the Outer Integral

Now evaluate the outer integral:\[ \int_{x = 0}^{\ln 2} e^x \, dx \]The integral of \( e^x \) with respect to \( x \) is still \( e^x \). Evaluate this from \( 0 \) to \( \ln 2 \).

Step 6: Evaluate the Result

Calculate the integral:\[ e^x \Big|_{0}^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1 \]Therefore, the area of the region is \( 1 \).

Key Concepts

Exponential Function

Exponential functions are mathematical functions of the form \( y = a e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. They are widely used in modelling growth processes, such as populations in biology, radioactive decay, and financial markets. The specific exponential function in our task is \( y = e^x \).

• This function passes through the point (0,1) because \( e^0 = 1 \).
• As \( x \) increases, \( y = e^x \) grows rapidly. This rapid increase is a hallmark of exponential growth.
• The graph is always above the x-axis because \( e^x \) is always positive.

Understanding the behavior of exponential functions is key in setting up and evaluating integrals involving areas under these curves.

Integration Limits

Integration limits define the boundaries over which we will integrate. They are crucial for computing the area under curves or within regions.For this exercise:- The integral is bounded in the \( x \) direction from \( x = 0 \) to \( x = \ln 2 \).- In the \( y \) direction, the bounds are from \( y = 0 \) to \( y = e^x \). This means for each fixed \( x \), \( y \) varies from 0 to the value of the exponential function at that \( x \).Choosing the correct integration limits is essential in accurately capturing the region of interest. Misplaced limits will result in incorrect area calculations.

Area Calculation

Calculating the area using a double integral involves evaluating the function within specified boundaries. Here's a breakdown of how it's done:1. **Setting up the Integral:** The problem involves setting up a double integral in the form \( \int_{x=0}^{\ln 2} \int_{y=0}^{e^x} dy \, dx \).2. **Evaluating the Inner Integral:** The inner integral, \( \int_{y = 0}^{e^x} dy \), is computed first. Since the integral of \( dy \) is \( y \), this evaluates to \( e^x \).3. **Evaluating the Outer Integral:** Next, compute the outer integral, \( \int_{x=0}^{\ln 2} e^x \, dx \). The integral of \( e^x \) with respect to \( x \) is \( e^x \).4. **Final Calculation:** Evaluate \( e^x \) from \( 0 \) to \( \ln 2 \), which results in \( e^{\ln 2} - e^0 = 2 - 1 = 1 \).Thus, the area of the defined region is 1. Such clear steps help ensure that calculations are systematic and accurate.