Problem 34 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{1}_{0} (5x - 5^x) \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{5}{2} - \frac{4}{\ln(5)}\).

Step 1: Split the Integral

The integral can be split into two parts: \( \int_0^1 5x \, dx \) and \( \int_0^1 5^x \, dx \). Thus, the expression becomes:\[\int_0^1 (5x - 5^x) \, dx = \int_0^1 5x \, dx - \int_0^1 5^x \, dx\]

Step 2: Evaluate \(\int_0^1 5x \, dx\)

To solve \(\int_0^1 5x \, dx\), use the power rule for integration. The antiderivative of \(5x\) is \(\frac{5x^2}{2}\). Calculate:\[\begin{align*}\int_0^1 5x \, dx & = \left. \frac{5x^2}{2} \right|_0^1 \& = \frac{5(1)^2}{2} - \frac{5(0)^2}{2} \& = \frac{5}{2}\end{align*}\]

Step 3: Evaluate \(\int_0^1 5^x \, dx\)

To solve \(\int_0^1 5^x \, dx\), use the formula for the integration of an exponential function, which gives the antiderivative \(\frac{5^x}{\ln(5)}\). Calculate:\[\begin{align*}\int_0^1 5^x \, dx & = \left. \frac{5^x}{\ln(5)} \right|_0^1 \& = \frac{5^1}{\ln(5)} - \frac{5^0}{\ln(5)} \& = \frac{5}{\ln(5)} - \frac{1}{\ln(5)} \& = \frac{4}{\ln(5)}\end{align*}\]

Step 4: Combine the Results

Subtract the result from Step 3 from the result in Step 2:\[\begin{align*}\int_0^1 (5x - 5^x) \, dx & = \frac{5}{2} - \frac{4}{\ln(5)}\end{align*}\]

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental tool you need to grasp when tackling integrals involving polynomial expressions. It is used to find the antiderivative, or the integral, of functions in the form of \( x^n \) where \( n eq -1 \). The rule states that:

• If \( f(x) = x^n \), then the antiderivative \( F(x) \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

Exploring this rule through our example, we are integrating the term \( 5x \). Here, \( n = 1 \), so we use the power rule to calculate:

• The antiderivative is \( \frac{5x^2}{2} \).

To evaluate this from 0 to 1, substitute the limits into the antiderivative and subtract:

• At \( x = 1 \): \( \frac{5(1)^2}{2} = \frac{5}{2} \).
• At \( x = 0 \): \( \frac{5(0)^2}{2} = 0 \).

Thus, the definite integral from 0 to 1 is \( \frac{5}{2} \). This shows how the power rule simplifies the process of finding integrals for polynomial terms.

Exponential Function Integration

Integration can become more complex when dealing with exponential functions like \( 5^x \). To evaluate integrals of the form \( a^x \) where \( a > 0 \), we utilize a specific rule for exponential functions. The antiderivative of \( a^x \) is:

• \( \frac{a^x}{\ln(a)} + C \), where \( \ln(a) \) is the natural logarithm of \( a \).

In our exercise, we are integrating \( 5^x \). By applying the rule for exponential functions, we find:

• The antiderivative is \( \frac{5^x}{\ln(5)} \).

We then evaluate this from 0 to 1:

• At \( x = 1 \): \( \frac{5^1}{\ln(5)} - \frac{5^0}{\ln(5)} = \frac{5 - 1}{\ln(5)} = \frac{4}{\ln(5)} \).

Integrals of exponential functions often involve natural logarithms, which is a key point to keep in mind when calculating such integrals. The examples show the necessity of understanding function properties to apply the correct rules.

Antiderivative Calculation

Calculating the antiderivative, sometimes called finding the "indefinite integral," is like finding the reverse of differentiation. For our exercise, we are dealing with definite integrals, meaning the result will be a number rather than a function with a \( +C \). Here are steps to follow for calculating an antiderivative:

• Identify the rule or formula applicable for the specific function.
• Apply the rule to find the antiderivative.
• Substitute the limits of integration into the antiderivative and perform the calculation.

In our task, we used two different rules:

• The power rule for the \( 5x \) term: Antiderivative is \( \frac{5x^2}{2} \), calculated from 0 to 1, totaling \( \frac{5}{2} \).
• The exponential rule for the \( 5^x \) term: Antiderivative is \( \frac{5^x}{\ln(5)} \), evaluated as \( \frac{4}{\ln(5)} \) from 0 to 1.

Finally, the result is calculated by combining the antiderivatives from the split terms. Understanding this process helps in evaluating integrals efficiently and accurately.