Problem 90 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int_{-2}^{0} 5^{-\theta} d \theta\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{24}{\ln(5)} \).

Step 1: Identify the function inside the integral

The given integral is \( \int_{-2}^{0} 5^{-\theta} \, d\theta \). This means we need to integrate the function \( 5^{-\theta} \) with respect to \( \theta \).

Step 2: Determine the antiderivative

To integrate \( 5^{-\theta} \), we need to use the formula for integrating exponential functions. The antiderivative of \( a^{-\theta} \) with respect to \( \theta \) is \( \frac{a^{-\theta}}{-\ln(a)} + C \), as long as \( a > 0 \) and \( a eq 1 \). Here, \( a = 5 \), so the antiderivative is \( \frac{5^{-\theta}}{-\ln(5)} \).

Step 3: Apply the limits of integration

We now evaluate the definite integral using the antiderivative: \[ \left[ \frac{5^{-\theta}}{-\ln(5)} \right]_{-2}^{0} = \frac{5^{0}}{-\ln(5)} - \frac{5^{-(-2)}}{-\ln(5)} \].

Step 4: Simplify the expression

Calculate \( 5^{0} \) which is 1, and \( 5^{2} \) which is 25. Substitute these into the expression from Step 3: \[ \frac{1}{-\ln(5)} - \frac{25}{-\ln(5)} = \frac{1 - 25}{-\ln(5)} = \frac{-24}{-\ln(5)} = \frac{24}{\ln(5)} \]. This is the value of the integral.

Key Concepts

Exponential Functions

Exponential functions are fundamental in mathematics, commonly denoted as \( a^x \), where \( a \) is a positive constant base and \( x \) is the exponent. They have distinct properties that make them incredibly useful, such as rapid growth or decay depending on the sign of the exponent.

Some key characteristics include:

• If \( a > 1 \), the function increases rapidly as \( x \) increases.
• If \( 0 < a < 1 \), the function decreases quickly as \( x \) increases.
• The graph of an exponential function has a horizontal asymptote typically at \( y=0 \).

In the context of integration, especially for negative exponents like \( 5^{-\theta} \), it involves understanding how these functions change over an interval. The base \( a \), here being 5, dictates the behavior of the function over specified bounds in the integral, such as from \( -2 \) to \( 0 \).

Antiderivative

The antiderivative, or the indefinite integral, is the reverse process of differentiation. For a function \( f(x) \), the antiderivative is a function \( F(x) \) such that \( F'(x) = f(x) \). It's noted by the integral symbol with no specified limits: \( \int f(x) \, dx \).

Calculating the antiderivative:

• For exponential functions like \( a^{-\theta} \), utilize the formula \( \frac{a^{-\theta}}{-\ln(a)} + C \).
• \( C \) is the constant of integration, symbolizing that indefinite integrals can have multiple solutions.

In our exercise, to find the antiderivative of \( 5^{-\theta} \), we applied this formula to find \( \frac{5^{-\theta}}{-\ln(5)} \), highlighting the necessity to understand and apply the formula correctly for definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with integration, providing a method for evaluating definite integrals. It states that if \( F \) is an antiderivative of \( f \) over an interval \([a, b]\), then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \].

Application steps include:

• Find the antiderivative \( F(x) \) of \( f(x) \).
• Evaluate \( F(x) \) at both bounds \( a \) and \( b \).
• Subtract the two results: \( F(b) - F(a) \).

In the given problem, after finding the antiderivative \( \frac{5^{-\theta}}{-\ln(5)} \), we applied this theorem. We calculated the difference between its values at the boundary points, \( 0 \) and \( -2 \), to find the definite integral: \[ \frac{24}{\ln(5)} \]. This demonstrates the power of this theorem in simplifying calculations and providing exact answers.