Chapter 8: Problem 25
Calculate. $$\int_{0}^{1} \ln \left(1+x^{2}\right) d x$$
Short Answer
Expert verified
The value of the integral \(\int_{0}^{1} \ln \left(1+x^{2}\right) dx\) is \(0\).
Step by step solution
01
Choose u and dv
Let's choose our u and dv as follows:
\[u = \ln \left(1+x^{2}\right)\]
and
\[dv = dx\]
02
Calculate du and v
Now we will differentiate u to get du and integrate dv to get v:
\[du = \frac{d (\ln \left(1+x^{2}\right))}{dx} = \frac{2x}{1+x^{2}}dx\]
and
\[v = \int dv = \int dx = x\]
03
Apply Integration by Parts
Using the integration by parts formula, we get:
\[\int_{0}^{1} \ln \left(1+x^{2}\right) dx = \left[x\ln \left(1+x^{2}\right)\right]_{0}^{1} - \int_{0}^{1} x \frac{2x}{1+x^{2}} dx\]
04
Simplify the Integral
Now let's simplify the integral:
\[\int_{0}^{1} \ln \left(1+x^{2}\right) dx = \left[\ln \left(1+x^{2}\right)\right]_{0}^{1} - 2\int_{0}^{1} \frac{x^2}{1+x^{2}} dx\]
05
Calculate the Remaining Integral
To deal with the remaining integral, we can substitute:
\[u = 1 + x^2\]
and
\[du = 2x dx\]
The integral becomes:
\[-2\int_{0}^{1} \frac{x^2}{1+x^{2}} dx = -\int_{1}^{2} \frac{du}{u}\]
Now, we can integrate:
\[-\int_{1}^{2} \frac{du}{u} = -[\ln |u|]_{1}^{2} = -(\ln 2 - \ln 1) = -\ln 2\]
06
Find the Final Answer
Now we put everything together to get the final answer:
\[\int_{0}^{1} \ln \left(1+x^{2}\right) dx = \left[\ln \left(1+x^{2}\right)\right]_{0}^{1} - 2\int_{0}^{1} \frac{x^2}{1+x^{2}} dx = (\ln 2 - \ln 1) - \ln 2 = 0\]
So, the value of the integral is:
\[\int_{0}^{1} \ln \left(1+x^{2}\right) dx = 0\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is a fundamental concept in calculus used to calculate the area under a curve within specific boundaries. In our case, we are dealing with the integral of \( \ln(1+x^2) \) from 0 to 1. This means we want to find the accumulated area between the function \( \ln(1+x^2) \), the x-axis, and the vertical lines x = 0 and x = 1.
Definite integrals provide a precise method to accumulate or "sum up" continuously varying quantities over a certain interval. The notation includes limits of integration, like \( \int_0^1 \), which define the start and end points. When solving a definite integral:
Definite integrals provide a precise method to accumulate or "sum up" continuously varying quantities over a certain interval. The notation includes limits of integration, like \( \int_0^1 \), which define the start and end points. When solving a definite integral:
- Evaluate the indefinite integral (anti-derivative) of the function.
- Apply the limits of integration by subtracting the value of the anti-derivative at the lower limit from its value at the upper limit.
- The result represents the "net area" between the curve and the x-axis.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions and have a natural base, which is e (Euler's number, approximately 2.718). A common logarithmic function is \( \ln(x) \), representing the logarithm with base e. These functions are essential in calculus and real-world applications, often appearing as solutions for differential equations and in growth processes.
In the exercise, the function \( \ln(1 + x^2) \) demonstrates a logarithm applied to a composition \((1 + x^2)\), which is typical in integration problems. When dealing with integrals involving logarithms:
In the exercise, the function \( \ln(1 + x^2) \) demonstrates a logarithm applied to a composition \((1 + x^2)\), which is typical in integration problems. When dealing with integrals involving logarithms:
- Differentiating \( \ln(x) \) yields \( \frac{1}{x} \).
- For the derivative of \( \ln(1 + x^2) \), apply the chain rule, which results in \( \frac{2x}{1+x^2} \).
- Logarithmic relationships help simplify problems, especially when used with properties such as \( \ln(a) - \ln(b) = \ln(\frac{a}{b}) \).
Substitution Method
The substitution method is a powerful tool used to simplify and evaluate integrals by changing variables. This method can make integration more manageable, especially when direct integration is difficult or impossible. The process involves selecting a part of the integral to substitute with a new variable, simplifying the integration process.
During substitution:
During substitution:
- Choose a substitution \( u = g(x) \) to simplify the expression. Ensure that the derivative \( du \) is present in the integrand.
- Replace the x terms with u terms, and express dx in terms of du.
- Adjust the limits of integration to the new variable, if dealing with definite integrals.