Problem 23 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{1}^{2} \frac{v^{3}+3 v^{6}}{v^{4}} d v\)

Step-by-Step Solution

Answer

\( \ln 2 + 7 \)

Step 1: Simplify the Integrand

First, simplify the integrand by dividing each term in the numerator by the denominator. The original integrand is \( \frac{v^3 + 3v^6}{v^4} \). Simplifying gives: \[ \frac{v^3}{v^4} + \frac{3v^6}{v^4} = v^{-1} + 3v^2. \]

Step 2: Set Up the Integral

Write the integral with the simplified integrand: \[ \int_{1}^{2} (v^{-1} + 3v^2) \, dv. \]

Step 3: Integrate Term by Term

Integrate each term separately: - The antiderivative of \( v^{-1} \) is \( \ln|v| \).- The antiderivative of \( 3v^2 \) is \( v^3 \).Thus, the antiderivative of the entire integrand is \( \ln|v| + v^3 \).

Step 4: Evaluate the Definite Integral

Apply the limits of integration, from 1 to 2: \[ \left[ \ln|v| + v^3 \right]_{1}^{2} = \left( \ln|2| + 2^3 \right) - \left( \ln|1| + 1^3 \right). \]

Step 5: Calculate the Result

Substitute and simplify the expression: - \( \ln|2| + 2^3 = \ln 2 + 8 \).- \( \ln|1| + 1^3 = 0 + 1 = 1 \). Then, the integral evaluates to: \[ \ln 2 + 8 - 1 = \ln 2 + 7. \]

Key Concepts

Integration Techniques

Evaluating definite integrals often requires applying a mix of integration techniques to simplify the problem. In this exercise, the integrand \( \frac{v^3 + 3v^6}{v^4} \) can initially seem challenging to tackle directly. However, by using algebraic manipulation, such as simplifying rational expressions, the task becomes more manageable.
By breaking down the expression, we use one common integration technique: simplifying first. This involves reducing the expression \( \frac{v^3}{v^4} + \frac{3v^6}{v^4} \) to \( v^{-1} + 3v^2 \).
This simplification step makes the application of standard antiderivatives easier in subsequent steps. Once simplified, the integral can be approached term by term, which is a basic technique of integrating each part separately before adding the results together.

Antiderivatives

Finding antiderivatives is central to solving integrals. An antiderivative, or the indefinite integral, is essentially the inverse operation of differentiation. With the integral \( \int (v^{-1} + 3v^2) \, dv \), identifying the antiderivative of each term is crucial.
Breaking down the process, the antiderivative of \( v^{-1} \) yields \( \ln|v| \), a common form when dealing with expressions involving the reciprocal of \( v \). For \( 3v^2 \), the antiderivative is \( v^3 \), since integrating constants multiplied by powers of \( v \) involves increasing the power by one and then dividing by the new power. This results in:

• \( v^{-1} \rightarrow \ln|v| \)
• \( 3v^2 \rightarrow v^3 \)

The summation of these results gives the composite antiderivative \( \ln|v| + v^3 \), which is evaluated over the given bounds.

Simplifying Rational Expressions

Simplifying rational expressions can greatly enhance the efficiency of solving integrals. In the problem \( \frac{v^3 + 3v^6}{v^4} \), each ratio of the polynomial terms can be simplified by separating the expression into simpler terms.
This involves performing division independently on each part of the fraction:

• For \( v^3/v^4 \), simplifying yields \( v^{-1} \).
• For \( 3v^6/v^4 \), it simplifies to \( 3v^2 \).

Through these steps, we avoid complex polynomial division, which can often lead to more cumbersome expressions. By focusing on lowering the powers, we reach a much simpler integrand of \( v^{-1} + 3v^2 \).
Simplifying before integration aids in revealing straightforward antiderivatives, thus facilitating quicker and more accurate calculations.