Problem 81 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{0}^{1} \frac{x}{x^{2}+1} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \ln 2 \).

Step 1: Understanding the Integral Structure

We need to evaluate the definite integral \[ \int_{0}^{1} \frac{x}{x^{2}+1} \, dx \] This integral involves the function \( \frac{x}{x^2 + 1} \). For integration, we might look for a method like substitution.

Step 2: Selecting a Substitution

To simplify the function \( \frac{x}{x^2+1} \), let us choose the substitution \( u = x^2 + 1 \). Then, the derivative \( du = 2x\, dx \) can be rewritten in terms of \( dx \) as \( dx = \frac{du}{2x} \).

Step 3: Substituting and Adjusting the Limits

Substituting \( u = x^2 + 1 \) and \( dx = \frac{du}{2x} \), the integral becomes:\[ \int_{x=0}^{x=1} \frac{x}{u} \cdot \frac{du}{2x} = \int_{u=1}^{u=2} \frac{1}{2u} \, du \] The limits of integration change since when \( x = 0 \), \( u = 0^2 + 1 = 1 \) and when \( x = 1 \), \( u = 1^2 + 1 = 2 \).

Step 4: Evaluate the New Integral

Now, evaluate the integral \[ \frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du \] This integrates to: \[ \frac{1}{2} [\ln |u|]_{1}^{2} \] which is: \[ \frac{1}{2} (\ln |2| - \ln |1|) \]

Step 5: Simplifying the Result

Since \( \ln |1| = 0 \), the expression simplifies to\[ \frac{1}{2} \ln 2 - \frac{1}{2} \cdot 0 = \frac{1}{2} \ln 2 \]Thus, the result of the definite integral is \( \frac{1}{2} \ln 2 \).

Key Concepts

Integration by Substitution

Integration by substitution is an effective technique to simplify complex integrals by introducing a new variable. This method works by changing the variable of integration to make the integrand easier to evaluate. To achieve this, we choose a substitution that relates the original variable to a new one. In the case of the integral \[ \int_{0}^{1} \frac{x}{x^2+1} \, dx \]we choose a substitution where \( u = x^2 + 1 \).

Why was this substitution chosen? Because the expression \( x^2 + 1 \) is part of the denominator, and its derivative, \( 2x \, dx \), closely matches the numerator, \( x \, dx \). This alignment simplifies the entire expression significantly when integrated.

When you substitute, always remember to:

• Calculate the differential for the new variable, here \( du = 2x \, dx \).
• Solve for the original \( dx \), leading to \( dx = \frac{du}{2x} \).
• Substitute both the variable and \( dx \) into the original integral.

These steps transform the integral into a simpler form that's often easier to evaluate.

Limits of Integration

In a definite integral, limits of integration define the interval over which the function is integrated. When applying substitution, adjusting the limits for the new variable is crucial since the original limits pertain to the original variable.

For our example, the integral \( \int_{0}^{1} \frac{x}{x^2+1} \, dx \)was initially set from \( x = 0 \) to \( x = 1 \). However, after the substitution \( u = x^2 + 1 \), these limits had to be recalculated in terms of \( u \).

To find the new limits:

• When \( x = 0 \): \( u = 0^2 + 1 = 1 \).
• When \( x = 1 \): \( u = 1^2 + 1 = 2 \).

Thus, the new limits for the integral become from \( u = 1 \) to \( u = 2 \). This ensures that after integration, the evaluation remains consistent with the original problem's boundaries.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in calculus, particularly when dealing with integrals. It is the inverse function of the exponential function with a base of \( e \), where \( e \approx 2.71828 \).

In our definite integral \[ \int_{0}^{1} \frac{x}{x^2+1} \, dx \] which becomes \[ \frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du \] after substitution, evaluating it results in \[ \frac{1}{2} [\ln |u|]_{1}^{2} \].

The integral of \( \frac{1}{u} \) results in a natural logarithm because the derivative of \( \ln u \) is \( \frac{1}{u} \). This makes it a go-to operation when the integrand matches this form.

• The resulting expression \[ \frac{1}{2} (\ln |2| - \ln |1|) \]
• Further simplifies to \( \frac{1}{2} \ln 2 \), as \( \ln 1 = 0 \).

Using natural logarithms in definite integrals highlights their usefulness and broad applicability in calculus.