Problem 53 - Applied Calculus

Question

For each definite integral: a. Evaluate it "by hand." b. Check your answer by using a graphing calculator. $$ \int_{0}^{1} \frac{x}{x^{2}+1} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Integration Method

To evaluate the integral \( \int_{0}^{1} \frac{x}{x^{2}+1} \, dx \), we will use substitution. This integral is suitable for substitution because the derivative of the denominator \( x^2 + 1 \) is present in the numerator (\( x \)).

Step 2: Perform Substitution

Let \( u = x^2 + 1 \), so \( du = 2x \, dx \). Therefore, \( \frac{1}{2} du = x \, dx \). Rewrite the integral in terms of \( u \):\[\int \frac{x}{x^2+1} \, dx = \int \frac{1}{2} \frac{1}{u} \, du = \frac{1}{2} \int \frac{1}{u} \, du.\]

Step 3: Evaluate the Integral

The integral \( \int \frac{1}{u} \, du \) is equal to \( \ln |u| + C \). Therefore:\[\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C.\]

Step 4: Substitute Back and Apply Limits

Substituting \( u = x^2 + 1 \) back into the expression gives:\[\frac{1}{2} \ln |x^2 + 1| igg|_0^1.\]Evaluate it by substituting the limits:\[\frac{1}{2} \ln(1^2 + 1) - \frac{1}{2} \ln(0^2 + 1) = \frac{1}{2} \ln(2) - \frac{1}{2} \ln(1).\]

Step 5: Simplify the Final Answer

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

Step 6: Verify Using a Graphing Calculator

Use a graphing calculator to verify the definite integral by entering it exactly: \( \int_0^1 \frac{x}{x^2+1} \, dx \). The calculated result should match \( \frac{1}{2} \ln(2) \approx 0.34657 \).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful method used to simplify complex integrals, making them easier to evaluate. In substitution, the idea is to convert the integral into a simpler form by changing the variable of integration. Here's how you might approach it:

• Identify a substitution that simplifies the integral. Typically, you look for a function in the integrand whose derivative is also present.
• In our exercise, we have the integral \( \int_{0}^{1} \frac{x}{x^2+1} \, dx\). Notice that the derivative of \( x^2 + 1 \) is \( 2x \), which resembles the numerator. This makes substitution feasible.
• Set \( u = x^2 + 1 \), then calculate \( du = 2x \, dx \), solving for \( x \, dx \), we find \( x \, dx = \frac{1}{2} du \).
• Rewrite the integral with these new terms: \( \int \frac{1}{2} \frac{1}{u} \, du \). This transformation simplifies the calculation significantly, since this now resembles a basic form of logarithmic integration.

By choosing an appropriate substitution, you maintain the integrity of the limits of integration from 0 to 1, ensuring continuity in solving definite integrals.

Graphing Calculator Verification

After solving a definite integral by hand, it's prudent to use a graphing calculator for verification. This step confirms your solution's accuracy and familiarity with technology-based tools enhances your mathematical toolkit.

• Enter the original integral expression \( \int_{0}^{1} \frac{x}{x^2+1} \, dx \) into the graphing calculator.
• The calculator should be set to calculate definite integrals over the specified interval \([0, 1]\).
• Check the numerical output. It should match the simplified answer from the hand-calculated method \( \frac{1}{2} \ln(2) \), which approximately equals 0.34657.

Using the graphing calculator not only proves your solution is correct but also builds confidence in utilizing technology for solving more complex integrals, where manual calculation might be cumbersome.

Logarithmic Integration

Logarithmic integration is a technique used when dealing with integrals where the integrand's form resembles a logarithmic derivative. The most common example is when facing integrals of the form \( \frac{1}{u} \), which integrate to natural logarithms.

• Once substitution is complete and you're left with \( \int \frac{1}{u} \, du \), the solution becomes \( \ln |u| + C \), where \( C \) is the constant of integration.
• In the exercise example, post-substitution results in evaluating \( \frac{1}{2} \ln |u| \) with the definite limits. Given our substitution \( u = x^2 + 1 \), the evaluated integral becomes \( \frac{1}{2} \ln |x^2 + 1| \bigg|_0^1 \).
• When evaluating definite integrals, consider the limits: substituting \( x = 1 \) and \( x = 0 \) gives \( \frac{1}{2} \ln(2) - \frac{1}{2} \ln(1) \). Since \( \ln(1) = 0 \), it simplifies to \( \frac{1}{2} \ln(2) \).

Understanding how to integrate logarithmically is essential for numerous problems, especially those involving rates of change and growth processes, where logarithmic behavior is prevalent.