Problem 121 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{1} \frac{1}{1+2 u} d u $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{1}{2} \ln 3 \).

Step 1: Identify the Integral Type

Recognize that the given integral is a definite integral of the form \( \int_{a}^{b} f(u) \, du \). In this case, \( f(u) = \frac{1}{1+2u} \), \( a = 0 \), and \( b = 1 \).

Step 2: Determine the Antiderivative

The function \( f(u) = \frac{1}{1+2u} \) resembles the derivative of a natural logarithm function. It can be rewritten as \( \frac{1}{2} \times \frac{1}{\frac{1}{2} + u} \). Thus, the antiderivative is \( \frac{1}{2} \ln |1+2u| + C \), where \( C \) is the constant of integration.

Step 3: Apply the Limits of Integration

Apply the limits of integration to the antiderivative. Calculate the difference: \[ \frac{1}{2} \ln |1+2(1)| - \frac{1}{2} \ln |1+2(0)| \] which simplifies to \[ \frac{1}{2} \ln 3 - \frac{1}{2} \ln 1 \].

Step 4: Simplify the Evaluated Expression

Since \( \ln 1 = 0 \), the expression simplifies further to: \[ \frac{1}{2} \ln 3 - 0 = \frac{1}{2} \ln 3. \]

Key Concepts

Antiderivative

When dealing with definite integrals, one crucial step is finding the antiderivative of the function. The antiderivative, also known as an indefinite integral, reverses the process of differentiation. Consider the original integrand \( f(u) = \frac{1}{1+2u} \). To determine its antiderivative, we first rewrite the expression as \( \frac{1}{2} \times \frac{1}{\frac{1}{2} + u} \). This expression hints at a natural logarithmic function, because the derivative of \( \ln(x) \) is \( \frac{1}{x} \).

• Rewriting the integrand helps recognize patterns of known derivatives.
• Using properties of logarithms and constants is often necessary to correctly identify the antiderivative.

Thus, the antiderivative of \( f(u) \) is \( \frac{1}{2} \ln |1+2u| + C \), where \( C \) is the constant of integration. This constant is important for indefinite integrals but does not affect the result of definite integrals, since the definite integral evaluates the change in the function over an interval.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm that has the constant \( e \) (approximately 2.71828) as its base. It's a critical function when working with integrals involving fractions like \( \frac{1}{x} \), due to its straightforward derivative.

• \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \)
• This property makes the natural logarithm valuable for integration.

In the context of our integral, recognizing \( \frac{1}{1+2u} \) resembles the derivative form of a natural logarithm is key. When we find that the antiderivative involves \( \ln |1+2u| \), we're utilizing the property that the derivative of a natural logarithm is a simple algebraic fraction. This connection allows us to rewrite complex fractions in a manageable logarithmic form, facilitating the calculation of definite integrals.

Limits of Integration

Once the antiderivative of a function is determined, applying the limits of integration is the next crucial step. These limits \( a \) and \( b \) correspond to the lower and upper bounds, respectively, in the definite integral notation \( \int_a^b f(x) \, dx \).

• The lower limit \( a \) and the upper limit \( b \) are substituted into the antiderivative.
• Calculation of the difference between these two evaluations results in the value of the definite integral.

For the integral \( \int_{0}^{1} \frac{1}{1+2u} \, du \), we use the antiderivative \( \frac{1}{2} \ln |1+2u| \). Substitute 1 and 0 into this antiderivative:\[ \frac{1}{2} \ln |1+2(1)| - \frac{1}{2} \ln |1+2(0)| \]This evaluates to \( \frac{1}{2} \ln 3 - 0 \), since \( \ln 1 = 0 \). Through this process, we transform the problem into a straightforward subtraction using the fundamental theorem of calculus.