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Magazine Chapter 5: Problem 53
For each definite integral: a. Evaluate it "by hand." b. Check your answer by using a graphing calculator. $$ \int_{0}^{1} \frac{x}{x^{2}+1} d x $$
Short Answer
Expert verified
The definite integral evaluates to \( \frac{1}{2} \ln(2) \).
Step by step solution
01
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 (\( x \)).
02
Perform Substitution
Let \( u = x^2 + 1 \), so \( du = 2x \, dx \). Therefore, \( \frac{1}{2} du = x \, dx \). Rewrite the integral in terms of \( u \):\[\int \frac{x}{x^2+1} \, dx = \int \frac{1}{2} \frac{1}{u} \, du = \frac{1}{2} \int \frac{1}{u} \, du.\]
03
Evaluate the Integral
The integral \( \int \frac{1}{u} \, du \) is equal to \( \ln |u| + C \). Therefore:\[\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C.\]
04
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:\[\frac{1}{2} \ln(1^2 + 1) - \frac{1}{2} \ln(0^2 + 1) = \frac{1}{2} \ln(2) - \frac{1}{2} \ln(1).\]
05
Simplify the Final Answer
Since \( \ln(1) = 0 \), the expression simplifies to:\[\frac{1}{2} \ln(2).\]
06
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 \( \frac{1}{2} \ln(2) \approx 0.34657 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 \( \frac{1}{2} \ln(2) \).
