Problem 45 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{0}^{\infty} \frac{d x}{(x+1)^{3 / 2}} $$

Step-by-Step Solution

Answer

The improper integral is convergent, and its value is 2.

Step 1: Set Up the Integral

Consider the given improper integral \( \int_{0}^{\infty} \frac{dx}{(x+1)^{3/2}} \). Since the integral is defined from 0 to infinity, it is improper, and we will need to evaluate its convergence and potential value using limits.

Step 2: Replace Infinity with a Variable

Rewrite the integral by replacing infinity with a variable \( t \), obtaining \( \int_{0}^{t} \frac{dx}{(x+1)^{3/2}} \). We will evaluate this integral and then take the limit as \( t \) approaches infinity.

Step 3: Evaluate the Integrated Function

Use substitution to evaluate the integral. Let \( u = x + 1 \), then \( du = dx \). The limits change accordingly: when \( x = 0, u = 1 \) and when \( x = t, u = t+1 \). This changes the integral to \( \int_{1}^{t+1} u^{-3/2} du \).

Step 4: Perform the Integration

Integrate \( u^{-3/2} \), whose antiderivative is \( -2u^{-1/2} \). This gives us \( -2u^{-1/2} \bigg|_{1}^{t+1} = -2(t+1)^{-1/2} + 2(1)^{-1/2} \).

Step 5: Apply the Limit

Take the limit as \( t \) approaches infinity: \[\lim_{t \to \infty} \left(-2(t+1)^{-1/2} + 2\right).\]The term \( -2(t+1)^{-1/2} \) approaches 0 as \( t \) increases, because \( (t+1)^{-1/2} \to 0 \), so the expression reduces to \( 2 \).

Step 6: Conclude Convergence and Value

Since the limit exists and is finite, the integral is convergent. The value of the integral is \( 2 \).

Key Concepts

Convergence

In calculus, determining whether an improper integral is convergent or divergent is crucial. When we say an integral is convergent, it simply means that the area under the curve defined by the integrand approaches a finite value as the limits of integration extend to infinity. Contrarily, a divergent integral implies the area becomes infinite, and thus the integral does not resolve to a specific value. To test convergence, we often replace the infinite limit with a variable and then take the limit of the integral as this variable approaches infinity. If this limit exists and is finite, the integral converges. For the integral \( \int_{0}^{\infty} \frac{dx}{(x+1)^{3/2}}\),substituting the upper limit with a variable \( t \) and evaluating it leads us to discover that this expression indeed converges, settling at a finite number which confirms its convergence.

Integration by Substitution

Integration by substitution is a powerful technique to simplify the integration process. It involves changing the variable in the integral to one that might make the integration easier. For the given exercise, we started with \( \int_{0}^{t} \frac{dx}{(x+1)^{3/2}}\).By letting \( u = x + 1 \), this transformed the integral into one with respect to \( u \):\( \int_{1}^{t+1} u^{-3/2} du.\)

• The main advantage of substitution is that it often converts complex expressions into simpler forms.
• It helps in finding the antiderivative, which is necessary to evaluate definite integrals.

This technique not only simplifies our work but also prepares the function to be assessed by limits, thereby aiding in determining convergence.

Limits in Calculus

Limits are foundational in calculus, especially when dealing with improper integrals. By evaluating the limit like \( \lim_{t \to \infty} \),we try to understand the behavior of a function as it approaches a particular value.In our exercise, after substituting and integrating, we have an expression of\(-2(t+1)^{-1/2} + 2\).

• As \( t \) approaches infinity, \((t+1)^{-1/2}\) approaches 0, simplifying our overall expression.
• This limit helps us determine the final value of the integral, thus confirming its convergence.

It's important to note that identifying when expressions tend toward 0, infinity, or a finite number is vital in studying the behavior of integrals and exploring their convergence or divergence.