Problem 101 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(93-106.\) $$\int_{0}^{2} \frac{\log _{2}(x+2)}{x+2} d x$$

Step-by-Step Solution

Answer

\(\frac{3}{2} \ln(2)\)

Step 1: Rewrite the Logarithm with Natural Logarithm

The given integral contains a logarithm with base 2, which can be transformed into a natural logarithm. Use the change of base formula: \[\log_2(y) = \frac{\ln(y)}{\ln(2)}\]to rewrite the integrand as \[\int_{0}^{2} \frac{\ln(x+2)}{(x+2)\ln(2)} \, dx.\]

Step 2: Simplify the Integral

Since \(\ln(2)\) is a constant, you can factor it out of the integral. This gives:\[\frac{1}{\ln(2)} \int_{0}^{2} \frac{\ln(x+2)}{x+2} \, dx.\]

Step 3: Use a Substitution Method

Let \(u = x + 2\). Then \(du = dx\). When \(x = 0\), \(u = 2\). When \(x = 2\), \(u = 4\). Substitute these into the integral, leading to:\[\frac{1}{\ln(2)} \int_{2}^{4} \frac{\ln(u)}{u} \, du.\]

Step 4: Recognize a Common Integral Form

The integral \(\int \frac{\ln(u)}{u} \, du\) is a standard form which equals \(\frac{1}{2}(\ln(u))^2 + C\), where \(C\) is a constant of integration.

Step 5: Evaluate the Definite Integral

Now, apply the definite integral from 2 to 4:\[\frac{1}{\ln(2)} \left[ \frac{1}{2}(\ln(4))^2 - \frac{1}{2}(\ln(2))^2 \right].\]Calculate the expression:\[\ln(4) = 2\ln(2)\text{, and so the integral becomes }\frac{1}{2}(2\ln(2))^2 = 2(\ln(2))^2.\]This means:\[\frac{1}{\ln(2)} \left[ 2(\ln(2))^2 - \frac{1}{2}(\ln(2))^2 \right] = \frac{1}{\ln(2)} \left[ \frac{3}{2}(\ln(2))^2 \right].\]

Step 6: Simplify the Final Result

Simplifying further:\[\frac{1}{\ln(2)} \times \frac{3}{2} \times (\ln(2))^2 = \frac{3}{2} \cdot \ln(2)\]Thus, the value of the integral is \(\frac{3}{2} \ln(2)\).

Key Concepts

Definite Integrals

Definite integrals are an essential part of integral calculus and are used to calculate the accumulated value of a function over a specific interval. In essence, when you work with definite integrals, you aim to find the area under the curve of a function between two endpoints on the x-axis. This is different from indefinite integrals, which provide a general form of the antiderivative.

With definite integrals, you substitute the upper and lower limits into the antiderivative to find the difference. The calculations are expressed as:

• \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\)

Where \(F(x)\) is the antiderivative of \(f(x)\).

In the exercise, the definite integral \(\int_{0}^{2} \frac{\log_2(x+2)}{x+2} \, dx\) is evaluated using limits from 0 to 2, representing a fixed area under the curve of the given function.

Logarithmic Integration

Logarithmic integration refers to solving integrals that involve logarithmic functions. These can be somewhat challenging due to the properties of logarithms, such as their ability to transform into natural logarithms using a change of base formula.

For a logarithmic integral like \(\int \frac{\ln(x+2)}{x+2} \, dx\), you'll often convert each logarithmic expression into its natural logarithm counterpart using:

• \(\log_b(y) = \frac{\ln(y)}{\ln(b)}\)

This helps to simplify integration processes, leading to more manageable expressions.

Once converted, the integration steps may include recognizing standard integral forms and proceeding with substitution methods if needed. This conversion process is vital in making the integration procedure smoother and more accessible, as seen since it turned our base 2 logarithm \(\log_2(x+2)\) into \(\frac{\ln(x+2)}{\ln(2)}\).

Integration Techniques

In calculus, integration techniques come in handy when dealing with more complex integrals. Among these techniques, substitution is quite common. It involves changing the variable of integration to simplify the integral's form, which was crucial in our example.

For the integral \(\int \frac{\ln(x+2)}{x+2} \, dx\), a simple substitution \(u = x + 2\) led to more straightforward limits and expression in terms of \(u\), becoming \(\int \frac{\ln(u)}{u} \, du\).

• This particular integral is in a standard form, \(\int \frac{\ln(u)}{u} \, du = \frac{1}{2}(\ln(u))^2 + C\), which is simpler to solve.

By applying definite limits after this transformation, we can reach the solved expression efficiently.

Efficient use of substitution helps in obtaining results faster by reducing the complexity of the given integrals, making it a valuable tool for students learning integral calculus.