Problem 1 - Calculus Single Variable

Question

In each of Exercises \(1-20\), determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{3}^{\infty} x^{-3 / 2} d x $$

Step-by-Step Solution

Answer

The integral converges and evaluates to \( \frac{2}{\sqrt{3}} \).

Step 1: Review the integral's form

The given integral is \( \int_{3}^{\infty} x^{-3/2} \, dx \). This is an improper integral with an infinite upper limit.

Step 2: Identify the type of integral

This integral is of the form \( \int_{a}^{\infty} x^n \, dx \), where \( n = -\frac{3}{2} \). We will use the convergence test for improper integrals with power functions, which states that the integral \( \int_{a}^{\infty} x^n \, dx \) converges if \( n < -1 \).

Step 3: Apply the convergence test

The exponent \( n = -\frac{3}{2} \) is less than \(-1\). Therefore, the integral \( \int_{3}^{\infty} x^{-3/2} \, dx \) converges.

Step 4: Evaluate the antiderivative

Find the antiderivative of \( x^{-3/2} \). The antiderivative is given by \( F(x) = \int x^{-3/2} \, dx = \frac{x^{-3/2+1}}{-3/2+1} + C = \frac{x^{-1/2}}{-1/2} = -2x^{-1/2} + C. \)

Step 5: Substitute limits and evaluate the definite integral

We need to evaluate \( \lim_{b \to \infty} \left[-2x^{-1/2}\right]_{3}^{b} \). This becomes \[ \lim_{b \to \infty} \left(-2b^{-1/2} - (-2 \cdot 3^{-1/2})\right) = \lim_{b \to \infty} -2b^{-1/2} + 2 \cdot \frac{1}{\sqrt{3}}. \] As \( b \to \infty \), \( -2b^{-1/2} \to 0 \). Thus, the result is \( 2 \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}. \)

Key Concepts

Convergence Test

When dealing with improper integrals, it is essential first to determine whether the integral converges or diverges. This is where the convergence test comes in handy, especially for power functions. For integrals of the form \( \int_{a}^{\infty} x^n \ dx \), the convergence or divergence depends on the value of \( n \). To use the convergence test:

• If \( n < -1 \), the integral converges. This indicates that as \( x \) approaches infinity, the function \( x^n \) decreases quickly enough for its overall area under the curve to be finite.
• If \( n \geq -1 \), the integral diverges. This happens because \( x^n \) does not decrease fast enough, and the area becomes infinite.

In our given problem, \( n = -\frac{3}{2} \). Since \(-\frac{3}{2} < -1\), we conclude that the integral converges. This is the critical first step before proceeding to evaluate the integral.

Antiderivative

Finding the antiderivative, or the indefinite integral, of a function is crucial for solving integrals. The antiderivative is essentially the reverse of taking a derivative. For the function \( x^{-3/2} \), we look for a function whose derivative gives us \( x^{-3/2} \).To find the antiderivative of \( x^{-3/2} \), we use the formula for power functions: \[F(x) = \int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Applying this to \( x^{-3/2} \):

• Increase the exponent by 1: \(-3/2 + 1 = -1/2\)
• Divide by the new exponent: \( \frac{x^{-1/2}}{-1/2} \)

Thus, the antiderivative is \(-2x^{-1/2} + C\). This step enables us to evaluate the definite integral by substituting the limits.

Power Function Integration

Power function integration is a method we use to evaluate integrals that involve expressions like \( x^n \). It's one of the basic integration techniques taught due to its applicability in many problems, especially when dealing with polynomials or similar functions.The concept involves:

• Increasing the exponent by 1 and dividing by this new exponent, specifically: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]
• Making sure \( n eq -1 \), because if \( n = -1 \), it turns into a logarithmic function: \( \ln|x| + C\).

In our case, we applied power function integration to the integral of \( x^{-3/2} \). By using its antiderivative \( -2x^{-1/2} \), it set the stage to evaluate the definite integral within infinite limits.

Infinite Limits

When we are evaluating improper integrals, particularly with infinite limits like \( \int_{3}^{\infty} x^{-3/2} \, dx \), it's crucial to understand how to handle these limits. The concept of infinite limits squarely deals with evaluating an integral where at least one limit goes to infinity.The process involves:

• Finding the antiderivative of the function concerned.
• Replacing the variable with a finite upper limit, usually represented by a symbol (e.g., \( b \)) that approaches infinity.
• Evaluating the difference between the limits, including taking the limit as \( b \to \infty \).
• In cases where infinity is involved, terms involving \( b \) as a denominator approach zero, simplifying to constants.

In our problem, we evaluated: \[\lim_{b \to \infty} \left(-2b^{-1/2} + 2 \cdot \frac{1}{\sqrt{3}}\right)\]As \( -2b^{-1/2} \to 0 \), we were left with a finite result of \( \frac{2}{\sqrt{3}} \), confirming the convergence.