Problem 1
Question
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{\infty} \frac{d x}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \(\frac{\pi}{2}\).
1Step 1: Understanding the Integral
The given integral is \(\int_{0}^{\infty} \frac{d x}{x^{2}+1}\). We are supposed to find the definite integral from 0 to \(\infty\). The integrand \(\frac{1}{x^2 + 1}\) is a standard form associated with the arctangent function.
2Step 2: Identify the Antiderivative
Recognize that the antiderivative of \(\frac{1}{x^2 + 1}\) is \(\arctan(x)\). This comes from the fact that the derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\).
3Step 3: Integrate from 0 to \(b\) and Find Limit as \(b \to \infty\)
Express the integral from 0 to \(\infty\) as a limit: \[\lim_{b \to \infty} \int_{0}^{b} \frac{d x}{x^{2}+1} = \lim_{b \to \infty} \left[ \arctan(x) \right]_{0}^{b}\]This gives us:\[\lim_{b \to \infty} \left( \arctan(b) - \arctan(0) \right)\]
4Step 4: Evaluate \(\arctan(b)\) and \(\arctan(0)\)
\(\arctan(0) = 0\) because the tangent of 0 is 0. As \(b\) approaches \(\infty\), \(\arctan(b)\) approaches \(\frac{\pi}{2}\) because the tangent function has an asymptote at \(\frac{\pi}{2}\).
5Step 5: Calculate the Limit
Substitute the values back into the limit expression:\[\lim_{b \to \infty} \left( \arctan(b) - \arctan(0) \right) = \frac{\pi}{2} - 0 = \frac{\pi}{2}\]Thus, the value of the integral is \(\frac{\pi}{2}\).
Key Concepts
AntiderivativeLimit EvaluationArctangent Function
Antiderivative
Finding the antiderivative is a crucial step in evaluating a definite integral. When we look at the integral \( \int \frac{1}{x^2+1} \, dx \), we need to determine an antiderivative, which is a function whose derivative gives us back the integrand. In this case, recognizing the function is key.
The integrand \( \frac{1}{x^2+1} \) hints at the derivative of the arctangent function \( \arctan(x) \), known from calculus. This is because the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \).
Thus, we can identify the antiderivative of \( \frac{1}{x^2+1} \) as \( \arctan(x) \). This antiderivative is vital for evaluating the definite integral in the subsequent steps.
The integrand \( \frac{1}{x^2+1} \) hints at the derivative of the arctangent function \( \arctan(x) \), known from calculus. This is because the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \).
Thus, we can identify the antiderivative of \( \frac{1}{x^2+1} \) as \( \arctan(x) \). This antiderivative is vital for evaluating the definite integral in the subsequent steps.
Limit Evaluation
In calculus, evaluating an integral over an infinite interval often involves taking limits. This process is key to handle integrals like \( \int_{0}^{\infty} \frac{d x}{x^2+1} \).
The original problem requires evaluating this definite integral from 0 to infinity. To tackle this, we reformulate the problem as a limit: \( \lim_{b \to \infty} \int_{0}^{b} \frac{d x}{x^2+1} \). Essentially, we first evaluate the integral up to a finite number \( b \), then observe the behavior as \( b \rightarrow \infty \).
Substituting the antiderivative from before, we compute \( \left[ \arctan(x) \right]_{0}^{b} \). Here, \( \arctan(b) \) at \( x=b \) minus \( \arctan(0) \) at \( x=0 \) simplifies to \( \arctan(b) - 0 \).
Letting \( b \rightarrow \infty \), we reach the limit expression \( \lim_{b \to \infty} ( \arctan(b) - 0 ) \), which leads us to the eventual result of the definite integral.
The original problem requires evaluating this definite integral from 0 to infinity. To tackle this, we reformulate the problem as a limit: \( \lim_{b \to \infty} \int_{0}^{b} \frac{d x}{x^2+1} \). Essentially, we first evaluate the integral up to a finite number \( b \), then observe the behavior as \( b \rightarrow \infty \).
Substituting the antiderivative from before, we compute \( \left[ \arctan(x) \right]_{0}^{b} \). Here, \( \arctan(b) \) at \( x=b \) minus \( \arctan(0) \) at \( x=0 \) simplifies to \( \arctan(b) - 0 \).
Letting \( b \rightarrow \infty \), we reach the limit expression \( \lim_{b \to \infty} ( \arctan(b) - 0 ) \), which leads us to the eventual result of the definite integral.
Arctangent Function
The arctangent function, denoted as \( \arctan(x) \), is integral in solving certain types of integrals. Understanding its limits is crucial in calculus, particularly with integrals involving infinite boundaries.
Specifically, \( \arctan(x) \) represents the inverse tangent function. Its range is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), and it describes the angle whose tangent is \( x \).
To solve \( \int_{0}^{\infty} \frac{d x}{x^2+1} \), we evaluate \( \arctan(b) \) as \( b \rightarrow \infty \). As \( x \) approaches infinity in \( \arctan(x) \), the function approaches a horizontal asymptote at \( \frac{\pi}{2} \). This is an important property that helps find limits for integrals that extend towards infinity.
Similarly, calculating \( \arctan(0) \) gives us 0, as the tangent of 0 is also 0. These values allow us to conclude that the integral evaluates to \( \frac{\pi}{2} \) when we consider the full range from 0 to infinity.
Specifically, \( \arctan(x) \) represents the inverse tangent function. Its range is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), and it describes the angle whose tangent is \( x \).
To solve \( \int_{0}^{\infty} \frac{d x}{x^2+1} \), we evaluate \( \arctan(b) \) as \( b \rightarrow \infty \). As \( x \) approaches infinity in \( \arctan(x) \), the function approaches a horizontal asymptote at \( \frac{\pi}{2} \). This is an important property that helps find limits for integrals that extend towards infinity.
Similarly, calculating \( \arctan(0) \) gives us 0, as the tangent of 0 is also 0. These values allow us to conclude that the integral evaluates to \( \frac{\pi}{2} \) when we consider the full range from 0 to infinity.
Other exercises in this chapter
Problem 1
Evaluate the integrals in Exercises \(1-28\). $$ \int \frac{d y}{\sqrt{9+y^{2}}} $$
View solution Problem 1
The instructions for the integrals in Exercises \(1-10\) have two parts, one for the Trapezoidal Rule and one for Simpson's Rule. . $$ \int_{1}^{2} x d x $$
View solution Problem 1
Use the table of integrals at the back of the book to evaluate the integrals. \(\int \frac{d x}{x \sqrt{x-3}}\)
View solution Problem 1
Evaluate the integrals in Exercises \(1-14\) $$ \int_{0}^{\pi / 2} \sin ^{5} x d x $$
View solution