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. $$ f(x)=\frac{\log _{2}(x)}{x} ; a=32, b=64 $$

Short answer

The area is approximately 3.8115.

Step-by-step solution

Understanding the Problem

We need to find the area under the curve of the function \( f(x) = \frac{\log_2(x)}{x} \) between the limits \( a = 32 \) and \( b = 64 \). This requires calculating the definite integral \( \int_{32}^{64} \frac{\log_2(x)}{x} \, dx \).

Change of Base for Logarithm

We start by converting the base of the logarithm to the natural logarithm to make integration easier. Since \( \log_2(x) = \frac{\ln(x)}{\ln(2)} \), the function becomes \( f(x) = \frac{\ln(x)}{x \cdot \ln(2)} \).

Setting Up the Integral

Rewrite the integral using the natural logarithm: \[ \int_{32}^{64} \frac{\ln(x)}{x \cdot \ln(2)} \, dx. \] Since \( \ln(2) \) is a constant, it can be factored out of the integral. This gives: \[ \frac{1}{\ln(2)} \int_{32}^{64} \frac{\ln(x)}{x} \, dx. \]

Finding the Antiderivative

The integral \( \int \frac{\ln(x)}{x} \, dx \) can be evaluated using the substitution \( u = \ln(x) \). Then \( du = \frac{1}{x} \, dx \), which results in the integral \( \int u \, du \). The antiderivative of \( u \) is \( \frac{u^2}{2} \). Back-substituting gives us \( \frac{(\ln(x))^2}{2} \).

Calculating the Definite Integral

Now evaluate \[ \frac{1}{\ln(2)} \left[ \frac{(\ln(64))^2}{2} - \frac{(\ln(32))^2}{2} \right]. \] Calculate \( \ln(64) = 6 \ln(2) \) and \( \ln(32) = 5 \ln(2) \). Substituting these into the formula gives us: \[ \frac{1}{2 \ln(2)} \left[ (6 \ln(2))^2 - (5 \ln(2))^2 \right]. \] This simplifies to: \[ \frac{1}{2 \ln(2)} (36 \ln^2(2) - 25 \ln^2(2)) = \frac{11 \ln^2(2)}{2 \ln(2)}. \] Finally, cancel one \( \ln(2) \) from the numerator and denominator, resulting in \( \frac{11 \ln(2)}{2} \).

Final Calculation and Result

Finally, compute the numerical value: \( \ln(2) \approx 0.693 \). Therefore, the area under the graph is approximately \( \frac{11 \times 0.693}{2} = 3.8115 \).

Key concepts

Area Under the Curve

Finding the area under the curve of a function between two points is a common application of integrals in calculus. This area can be thought of as the accumulated quantity between the curve and the x-axis from one endpoint to another. It is especially important when dealing with functions of real-valued variables.
To compute the area under a curve, we use the definite integral. Here, the function given is \( f(x) = \frac{\log_2(x)}{x} \), and the interval is from \( a = 32 \) to \( b = 64 \). The integral from \( a \) to \( b \) will give the exact area under the curve, providing insight into how the function behaves across the specified interval.
This is particularly useful in physics and engineering, where such calculations might represent real-world quantities like distance, volume, or accumulated change.

Logarithm Base Change

In this problem, the original function involves a base 2 logarithm: \( \log_2(x) \). However, working directly with logarithms that are not natural might complicate integration. To ease the process, we employ the change of base formula. The formula \( \log_2(x) = \frac{\ln(x)}{\ln(2)} \) helps convert the expression into one involving the natural logarithm, \( \ln \).
By converting to natural logs, we have \( f(x) = \frac{\ln(x)}{x \cdot \ln(2)} \). Since \( \ln(2) \) is a constant, it doesn't affect the integral's structure significantly beyond a multiplicative factor.
This step simplifies further calculations because the natural logarithm has well-established integration techniques, unlike base 2 logarithms.

Substitution Method

Substitution is a powerful method that simplifies integration by changing variables. In the integral \( \int \frac{\ln(x)}{x} \, dx \), recognizing that \( \ln(x) \) seems like a part of an expression that can be substituted, leads us to define a new variable \( u = \ln(x) \).
Differentiating \( u \) with respect to \( x \), gives \( du = \frac{1}{x} \, dx \). This conveniently transforms the integral into \( \int u \, du \), which is much simpler to evaluate.
Substitution essentially breaks down complex integrals into more manageable parts, allowing you to tackle harder integrals using known methods.

Antiderivative Calculation

Calculating an antiderivative is identifying a function whose derivative is the given function. After substitution, we simplify \( \int u \, du \) to find the antiderivative of \( u \), which is \( \frac{u^2}{2} \).
By substituting back \( u = \ln(x) \), the expression becomes \( \frac{(\ln(x))^2}{2} \). This represents the antiderivative of our transformed integral.
To find the definite integral from \( a = 32 \) to \( b = 64 \), we substitute these values into \( \frac{(\ln(x))^2}{2} \) and compute the difference: \( \left[ \frac{(\ln(64))^2}{2} - \frac{(\ln(32))^2}{2} \right] \).
The final step involves multiplying by the constant \( \frac{1}{\ln(2)} \) and simplifying to arrive at the calculated area under the curve. This ends up as \( \frac{11 \ln(2)}{2} \), showcasing how antiderivatives lead to definite integrals and area calculations.