Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=e^{2 x}, y=0\), between \(x=0\) and \(x=\ln 2\)

Step-by-Step Solution

Answer

The area of the region is \(\frac{3}{2}\) or 1.5.

Step 1: Identifying the Region

We need to sketch the region bounded by the equations \(y = e^{2x}\) and \(y = 0\) between \(x = 0\) and \(x = \ln 2\). The function \(y = e^{2x}\) is an exponential curve that increases as \(x\) increases, and \(y = 0\) is the x-axis. The region of interest lies between the x-axis and the curve from \(x = 0\) to \(x = \ln 2\).

Step 2: Drawing the Typical Slice

A typical slice is a vertical strip between the graph of the function and the x-axis. For this problem, a slice would extend from the x-axis (\(y=0\)) to the curve \(y=e^{2x}\), with a small width \(\Delta x\).

Step 3: Setting Up the Integral

To find the area, we integrate the function \(y = e^{2x}\) with respect to \(x\) from \(x = 0\) to \(x = \ln 2\). The integral is set up as follows: \[A = \int_{0}^{\ln 2} e^{2x} \, dx\]

Step 4: Calculating the Integral

To calculate the integral \( \int e^{2x} \, dx \), we use substitution. Let \(u = 2x\), which gives \(du = 2dx\) or \(dx = \frac{du}{2}\). The limits of integration change accordingly: when \(x = 0\), \(u = 0\), and when \(x = \ln 2\), \(u = 2\ln 2 = \ln 4\). Rewriting the integral:\[A = \frac{1}{2} \int_{0}^{\ln 4} e^u \, du\]This evaluates to \(\frac{1}{2} [e^u]_0^{\ln 4}\). Calculate this:\[A = \frac{1}{2} (e^{\ln 4} - e^0) = \frac{1}{2} (4 - 1) = \frac{3}{2}\]

Step 5: Estimating the Area to Confirm Answer

To estimate, note that \(e^{2x}\) approximately equals 1 when \(x\) is near 0 and grows to 4 when \(x = \ln 2\). The average value over this interval can be approximated as roughly halfway between these values, giving an estimate of about \(2.5\). Multiplied by \(\ln 2 \approx 0.693\), we also get an approximate area of around 1.7. The calculated precise area of 1.5 is consistent with this rough estimate.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in integral calculus. It is used to calculate the accumulated quantity, like area under a curve, over a specific interval. When you solve a definite integral, you are essentially finding the net area between a given function and the x-axis over a finite interval.
The definite integral is represented as \( \int_{a}^{b} f(x) \, dx\), where \( f(x)\) is the function, and \(a\) and \(b\) are the limits of integration. These limits define the interval in which you're interested.

• The lower limit, \(a\), represents the starting point of the interval.
• The upper limit, \(b\), represents the endpoint of the interval.

This process calculates the total area between the function and the x-axis, considering both the areas above and below the x-axis. Positive area counts when the function is above the axis and negative when below.
Definite integrals transform the abstract notion of continuous accumulation into concrete values, such as finding the area under \( y = e^{2x}\) from \(x = 0\) to \(x = \ln 2\). Understanding it as a complete net area helps students analyze dynamic systems where accumulation over time or distance is crucial.

Area Under a Curve

The area under a curve in an interval is a vital concept in calculus that shows how a function behaves over that interval. Calculating this area helps in various real-world applications, like determining the distance traveled over time or creating economic forecasts.
To understand the area under a curve, envision slicing the area into narrow vertical strips. Each strip has:

• A very small base width, denoted as \( \Delta x\)
• A height determined by the function's value at that point, \( f(x)\)

The total area is approximated by summing the area of these strips, \( f(x) \cdot \Delta x\), which becomes precise as strip widths approach zero, leading to integration.
The integral \( \int_{0}^{\ln 2} e^{2x} \, dx\) computes the exact area formed between the curve \( y = e^{2x}\) and the x-axis from \( x = 0\) to \( x = \ln 2\). Seeing this as gathering many small parts into a whole is a pivotal analytical advance, boosting comprehension and application of mathematical growth patterns.

Exponential Functions

Exponential functions are crucial in calculus due to their distinct growth patterns and widespread applications across sciences and engineering. An exponential function typically has the form \( y = a \, e^{bx}\), where \( a\) and \( b\) are constants and \( e\) denotes Euler's number, approximately 2.718.
Key characteristics include:

• Rapid increase (or decrease) depending on the sign of \( b\).
• The function crosses the y-axis at \( y = a\) when \( x = 0\).
• An inherent natural growth factor represented by \( e\).

In the problem, \( y = e^{2x}\) grows quickly, creating a pronounced upward bend. This nature severely impacts the net area calculation under the curve, reflected in integrals, and aids in dynamic modeling. Through substitution, the function's integral simplifies, affirming its flexibility.
Whether modeling population growth or radioactive decay, understanding these functions fosters intuitive grasp of naturally occurring exponential changes. The example emphasizes both the mathematical elegance and practical importance of exponential curves in calculus and beyond.