Problem 10 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int p^{5} \ln p d p\)

Step-by-Step Solution

Answer

\(\frac{p^6}{6} \ln p - \frac{p^6}{36} + C\)

Step 1: Identify the Integration Technique

To solve the integral \(\int p^5 \ln p \, dp\), we notice that it involves both a polynomial and a logarithmic function. This is a perfect candidate for integration by parts, which is suitable for integrals of the form \(\int u \, dv\).

Step 2: Choose \(u\) and \(dv\)

For integration by parts, recall the formula \(\int u \, dv = uv - \int v \, du\). We choose \(u = \ln p\) and \(dv = p^5 \, dp\). This choice is made because differentiating \(\ln p\) becomes simpler, while \(p^5\) is still easily integrable.

Step 3: Compute \(du\) and \(v\)

Differentiate \(u = \ln p\) to get \(du = \frac{1}{p} \, dp\). Integrate \(dv = p^5 \, dp\) to get \(v = \frac{p^6}{6}\).

Step 4: Substitute into the Integration by Parts Formula

Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula: \[ \int u \, dv = uv - \int v \, du\]This gives: \[ \int p^5 \ln p \, dp = \frac{p^6}{6} \ln p - \int \frac{p^6}{6} \cdot \frac{1}{p} \, dp\] simplifying further: \[ = \frac{p^6}{6} \ln p - \frac{1}{6} \int p^5 \, dp\].

Step 5: Evaluate the Remaining Integral

Now, solve \(\int p^5 \, dp\). This is a straightforward power rule integration: \[ \int p^5 \, dp = \frac{p^6}{6} \]Substitute this back to get: \(\frac{p^6}{6} \ln p - \frac{1}{6} \cdot \frac{p^6}{6} + C\) where \(C\) is the constant of integration.

Step 6: Simplify the Expression

Simplify the expression obtained: \[ \frac{p^6}{6} \ln p - \frac{p^6}{36} + C \]This is the final solution, and it represents the antiderivative of the original integral.

Key Concepts

Polynomial Integration

Polynomial integration involves finding the antiderivative of polynomial expressions. This often uses the power rule, which states that for any polynomial term of the form \(x^n\), the integral is given by \(\int x^n \ dx = \frac{x^{n+1}}{n+1} + C\). Here, \(C\) represents the integration constant.

• For example, integrating \(x^5\) would result in \(\frac{x^6}{6} + C\).
• This rule is applied individually to each term in a polynomial expression when integrating.

In our specific exercise, once the logarithmic part has been addressed, we encounter a pure polynomial integral, \(\int p^5 \ dp\). Using the power rule, this simplifies directly to \(\frac{p^6}{6}\). Polynomial integrations are straightforward, as long as the constants are correctly identified and included in the result.

Logarithmic Integration

Logarithmic integration comes into play when an integral includes logarithmic functions such as \(\ln(x)\). These can complicate direct integration due to their nature. Often, these integrals are tackled using an integration technique called integration by parts.

• In our example, \(\ln(p)\) is part of the integrand, making it not straightforward to integrate directly. Hence, integration by parts is used, where \(\ln(p)\) is assigned as \(u\), the part we differentiate.
• This choice simplifies the problem, since differentiating \(\ln(p)\) leads to \(\frac{1}{p}\), a more manageable expression.

Understanding logarithmic integration is essential when encountering integrals where logarithmic terms like \(\ln(x)\) make direct integration difficult. The use of integration by parts plays a crucial role here, helping transform the integral into a solvable expression.

Antiderivative

An antiderivative is essentially the reverse of differentiation, particularly the process of finding a function whose derivative results in the original function. In integral calculus, when we find an antiderivative, it's often accompanied by a constant represented by \(C\). This is because any constant disappears when differentiating, making it part of an infinite set of possible solutions.

• In our example, solving for the integral ultimately gives us the antiderivative \(\frac{p^6}{6} \ln p - \frac{p^6}{36} + C\).
• Each term of the integral contributes to forming this expression, and the constant \(C\) reflects the family of functions that all satisfy the original integral equation.

Antiderivatives are foundational in calculus for understanding how functions accumulate values, crucial for applications in various physical and mathematical problems. Recognizing and calculating antiderivatives also prepares us for more advanced integration techniques. They provide insights into the accumulation and total impact, as seen in solving integrals like \(\int p^5 \ln p \ dp\).