Problem 135 - Thomas Calculus

Question

Find the area of the "triangular" region in the first quadrant that is bounded above by the curve \(y=e^{2 x}\) , below by the curve \(y=e^{x}\) and on the right by the line \(x=\ln 3 .\)

Step-by-Step Solution

Answer

The area is 2 square units.

Step 1: Set Up the Integral to Find the Area

To find the area between the curves, we first need to identify the region of integration. Since we are given the curves \(y = e^{2x}\) and \(y = e^x\), and the bounding line \(x = \ln 3\), we set up the integral from \(x = 0\) (where the curves intersect on the y-axis) to \(x = \ln 3\). The area \(A\) is given by the integral of the difference between the top curve and the bottom curve over this interval: \[ A = \int_{0}^{\ln 3} (e^{2x} - e^x) \, dx \].

Step 2: Integrate the Function

Now, integrate the function \(e^{2x} - e^x\). The integral of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\), and the integral of \(e^x\) is \(e^x\). Therefore, the integral becomes: \[ \int_{0}^{\ln 3} (e^{2x} - e^x) \, dx = \left[ \frac{1}{2}e^{2x} - e^x \right]_{0}^{\ln 3} \].

Step 3: Evaluate the Definite Integral

Substitute the upper limit \(x = \ln 3\) into the antiderivative: \[ \frac{1}{2}e^{2\ln 3} - e^{\ln 3} = \frac{1}{2}(e^{\ln 9}) - 3 = \frac{1}{2}(9) - 3 = 4.5 - 3 = 1.5 \]. Next, substitute the lower limit \(x = 0\) into the antiderivative: \[ \frac{1}{2}e^{0} - e^{0} = \frac{1}{2}(1) - 1 = 0.5 - 1 = -0.5 \].

Step 4: Calculate the Area

Subtract the evaluated lower limit from the upper limit: \[ A = (1.5) - (-0.5) = 1.5 + 0.5 = 2 \]. Thus, the area of the region is 2 square units.

Key Concepts

Definite Integral

In mathematics, a definite integral is used to calculate the area under a curve over a specified interval. The term 'definite' indicates that the integral is evaluated over a set interval, producing a number rather than another function.
This concept is crucial when determining the area between curves. To apply it, you need to know the bounds of the region you are interested in, which are the 'limits of integration'. In the case of finding the area between curves, the definite integral accounts for the variations in height between the top curve and the bottom curve across the entire span from the lower bound to the upper bound.

• The integral symbol (∫) denotes summation over an interval.
• Limits of the integration denote the start and end points. In our exercise example, from 0 to \( \ln 3 \).
• The expression inside the integral represents the function whose area is to be calculated. In our scenario: \( e^{2x} - e^x \).

By evaluating the definite integral, one finds the net "signed area" between the curves, which may include subtracting a portion if part of the area lies below the x-axis before the solutions are added together.

Exponential Functions

Exponential functions involve expressions where variables appear as exponents, such as \( e^{x} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.
Exponential functions are common in models for growth processes, like population or radioactive decay, making this concept essential in applied mathematics and sciences. In graphical terms, exponential functions are curves that generally show a sharp increase or decrease, depending on whether they are growing or decaying exponentially.
Understanding how to manipulate these functions is key to successfully solving integrals involving them. For example, in this exercise:

• The function \( e^{2x} \) grows more quickly than \( e^x \) since it is an exponential expression raised to a larger power.
• Such functions can be integrated using basic calculus rules, which are essential for finding areas between curves as demonstrated.
• Identifying the hierarchy of these functions helps set up integrals accurately, sorting which function is considered 'on top' or 'below' in a given interval.

Integration Techniques

Integration techniques are the methods used to solve integrals, which are necessary for finding the area between curves. The basic technique used in solving the given exercise is called 'basic integration', focusing on well-known forms like the exponential functions, \( e^x \) and \( e^{2x} \).
Here's a mini-guide to understanding basic techniques:

• The integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} \). In this scenario, the integral of \( e^{2x} \) is \( \frac{1}{2} e^{2x} \).
• The subtraction and addition of integrals come into play while dealing with multiple terms in functions, determining the net area as shown by \( (e^{2x} - e^x) \).
• After determining the antiderivatives, one needs to plug in the specified bounds to compute the definite integral of the function.

Each step from setting up to evaluating the integral applies these techniques to calculate the exact area, showcasing how integration not only helps us solve such mathematical problems but provides insights into real-world applications.