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. $$ f(x)=2^{-x} ; a=1, b=2 $$

Short answer

The area is \( \frac{1}{4\ln(2)} \).

Step-by-step solution

Understand the Problem

We are given a function, \( f(x) = 2^{-x} \), and we need to find the area under this function between \( x = 1 \) and \( x = 2 \). This requires us to integrate \( f(x) \) with respect to \( x \) from \( a = 1 \) to \( b = 2 \).

Set up the Integral

The area under \( f(x) = 2^{-x} \) from \( x = 1 \) to \( x = 2 \) can be found using the definite integral:\[ \int_{1}^{2} 2^{-x} \, dx \]

Find the Indefinite Integral

To solve \( \int 2^{-x} \, dx \), first note that the antiderivative of \( 2^{-x} \) involves the natural logarithm. We have the formula for exponential decay:\[ \int 2^{-x} \, dx = -\frac{2^{-x}}{\ln(2)} + C \] where \( C \) is the constant of integration.

Evaluate the Definite Integral

Substitute \( x = 2 \) and \( x = 1 \) into the antiderivative and find the definite integral:\[ \left[ -\frac{2^{-x}}{\ln(2)} \right]_{1}^{2} = \left(-\frac{2^{-2}}{\ln(2)}\right) - \left(-\frac{2^{-1}}{\ln(2)}\right) \] Simplify this expression:\[ = -\frac{1}{4\ln(2)} + \frac{1}{2\ln(2)} \]

Simplify the Expression

Combine the terms:\[ = \frac{1}{2\ln(2)} - \frac{1}{4\ln(2)} \] This simplifies to:\[ \frac{1}{4\ln(2)} \] Thus, the area under the curve from \( x = 1 \) to \( x = 2 \) is \( \frac{1}{4\ln(2)} \).

Key concepts

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a^x \), where \( a \) is a positive constant known as the base. In our exercise, the function \( f(x) = 2^{-x} \) is a specific type of exponential function that represents exponential decay. This means as \( x \) increases, the function value decreases. Exponential functions are unique because their rates of change are proportional to their function values.

Key Properties of Exponential Functions:

• They exhibit rapid growth or decay, depending on the base value.
• The base \( a \) is crucial: if \( a > 1 \), we have growth, and if \( 0 < a < 1 \), we have decay.
• They have continuous and smooth curves.

For \( f(x) = 2^{-x} \), this describes a decay because as \( x \) gets larger, \( 2^{-x} \) becomes smaller. Understanding this behavior helps in visualizing how the area under the curve behaves when integrated between a range of \( x \) values.

Antiderivative

An antiderivative is a function whose derivative is the original function. It is essentially the reverse process of differentiation. In integral calculus, finding an antiderivative is crucial for solving definite integrals, which help us determine the area under a curve.

To find the antiderivative of \( 2^{-x} \), we rely on a specific formula for exponential decay. The antiderivative is given by:\[\int 2^{-x} \, dx = -\frac{2^{-x}}{\ln(2)} + C\]Here, \( C \) represents the constant of integration, which is not necessary for definite integrals but is important for understanding the family of possible antiderivatives.

Key Points:

• The process of finding an antiderivative involves integrating the function.
• For exponential functions, the natural logarithm often appears in the result due to the properties of logarithms and exponentials.
• When evaluating a definite integral, the constant \( C \) drops out because we evaluate the antiderivative at the boundaries and subtract the results.

Numerical Integration

Numerical integration is a technique used to approximate the value of integrals, especially when analytical solutions are difficult or impossible to obtain. In some cases, functions do not have simple antiderivatives, and this is where numerical methods become handy, allowing for practical calculation of area under a curve.

Common Numerical Methods:

• **Trapezoidal Rule:** Approximates the area under the curve using trapezoids.
• **Simpson’s Rule:** Uses parabolic arcs instead of line segments for a more accurate approximation.
• **Midpoint Rule:** Uses the midpoint of intervals to approximate the area.

Though in our exercise, we found an analytical solution, numerical integration offers a robust tool for handling more complex functions. Understanding these methods can be very useful if exact solutions are cumbersome or impossible to obtain. In practice, computational tools and software often employ these techniques to handle real-world data and models.