Question

In the following exercises, f(x)?0 for a?x?b. Find the area under the graph of f(x) between the given values a and b by integrating. $$ f(x)=2^{-x} ; a=3, b=4 $$

Short answer

The area is \( \frac{1}{16 \ln(2)} \).

Step-by-step solution

Set up the integral

To find the area under the graph of the function \( f(x) = 2^{-x} \) from \( a = 3 \) to \( b = 4 \), we need to set up the definite integral: \[ \int_{3}^{4} 2^{-x} \, dx \] This integral represents the area under the curve between \( x = 3 \) and \( x = 4 \).

Find the antiderivative

Calculate the antiderivative of \( f(x) = 2^{-x} \). The antiderivative of \( 2^{-x} \) can be found using the formula for the integration of exponential functions:\[ \int 2^{-x} \, dx = -\frac{2^{-x}}{\ln(2)} + C \] where \( C \) is the constant of integration. However, since we are computing a definite integral, \( C \) will not appear in the final answer.

Evaluate the definite integral

Now substitute the limits into the antiderivative:\[ \left[ -\frac{2^{-x}}{\ln(2)} \right]_{3}^{4} = \left( -\frac{2^{-4}}{\ln(2)} - \left( -\frac{2^{-3}}{\ln(2)} \right) \right) \]Simplify this expression:\[ -\frac{1}{16 \ln(2)} + \frac{1}{8 \ln(2)} \]

Simplify the result

Combine the fractions:\[ \frac{1}{8 \ln(2)} - \frac{1}{16 \ln(2)} = \frac{2}{16 \ln(2)} - \frac{1}{16 \ln(2)} = \frac{1}{16 \ln(2)} \]

Finalize the solution

The area under the curve \( f(x) = 2^{-x} \) from \( x = 3 \) to \( x = 4 \) is \[ \frac{1}{16 \ln(2)} \].

Key concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In the context of the exercise we're considering, the function given is \( f(x) = 2^{-x} \). Here, 2 is the base, and the exponent is the variable \( -x \). These functions have unique properties that make them especially useful in modeling situations involving growth and decay.
For instance, a function with a positive exponent might model population growth, while a negative exponent, like in \( 2^{-x} \), represents decay or reduction. This specific function decreases over time as \( x \) increases. The curve of this function is smooth and gentle, sloping downwards as you move left-to-right along the x-axis.

• Exponential decay functions, like \( 2^{-x} \), are crucial for understanding natural processes such as radioactive decay or cooling processes.
• The base \( 2 \) indicates how steep or gradual the curve will be.

Antiderivative

The antiderivative, also known as the indefinite integral, is the process of finding a function whose derivative is the given function. For our purpose, finding the antiderivative of \( 2^{-x} \) involves utilizing the properties of exponential functions in calculus.
The general idea is to reverse the process of differentiation. When we integrate \( 2^{-x} \), we need to consider the integral rule for exponential functions. The integral of \( 2^{-x} \) with respect to \( x \) yields:

• \( \int 2^{-x} \, dx = -\frac{2^{-x}}{\ln(2)} + C \)

This general formula involves a constant \( C \), known as the constant of integration, but typically gets omitted in definite integrals because it cancels out. The trick in finding the antiderivative requires recognizing the base of the exponential function and applying the corresponding integration rule.

Integration Techniques

Integration techniques allow us to calculate the area under a curve of a function between certain limits. In our example, we calculate the area under \( f(x) = 2^{-x} \) from \( x = 3 \) to \( x = 4 \).
To achieve this, we must:

• Set up the definite integral as \( \int_{3}^{4} 2^{-x} \, dx \).
• Find the antiderivative of the function, which gives us \( -\frac{2^{-x}}{\ln(2)} \).
• Evaluate the integral by plugging in the upper and lower limits into the antiderivative.

Thus, choose the correct method tailored to the type of function you are dealing with, ensuring accuracy in your results. Calculating definite integrals requires substitution and simplification to arrive at the final result, which represents the area.

Area Under a Curve

The area under a curve between two points on a graph is a fundamental concept in calculus. It quantifies the total space between the curve and the x-axis over an interval. For \( f(x) = 2^{-x} \), calculating the area from \( x = 3 \) to \( x = 4 \) involves solving a definite integral.
This area represents the accumulated value of the function over that interval. The steps include:

• Finding the antiderivative of the function.
• Substituting the limits into the antiderivative expression.
• Simplifying to find the numerical value.

The result \( \frac{1}{16 \ln(2)} \) is the exact area under the curve from \( x = 3 \) to \( x = 4 \), providing insight into the behavior of the function within this specific range.