Problem 51 - Applied Calculus

Question

AREA Find the area between the curve \(y=1 / x^{3 / 2}\) and the \(x\) -axis from \(x=1\) to \(\infty\).

Step-by-Step Solution

Answer

The area is 2.

Step 1: Understand the Problem

We need to find the area between the curve given by the function \(y=\frac{1}{x^{3/2}}\) and the \(x\)-axis from \(x=1\) to \(\infty\). This involves evaluating an improper integral because the upper limit of integration is \(\infty\).

Step 2: Set up the Integral

To find the area, set up the improper integral \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \). This integral represents the area under the curve starting at \(x=1\) and extending to infinity.

Step 3: Integrate the Function

We need to evaluate the integral \( \int \frac{1}{x^{3/2}} \, dx \). First, rewrite the function: \( \frac{1}{x^{3/2}} = x^{-3/2} \). The antiderivative of \( x^{-3/2} \) is \( \frac{x^{-1/2}}{-1/2} = -2x^{-1/2} \). Thus, the integral becomes \( -2x^{-1/2} \).

Step 4: Evaluate the Improper Integral

Because the integral is improper, evaluate it as a limit: \[ \lim_{{b \to \infty}} \left( \int_{1}^{b} \frac{1}{x^{3/2}} \, dx \right) = \lim_{{b \to \infty}} \left( -2x^{-1/2} \right) \]. Evaluate from \(x=1\) to \(x=b\).

Step 5: Calculate the Limit

Substitute the bounds into the antiderivative: \[ \lim_{{b \to \infty}} \left( -2b^{-1/2} - (-2(1)^{-1/2}) \right) \]. Simplifying gives \[ \lim_{{b \to \infty}} (-2b^{-1/2} + 2) \]. As \(b\) approaches \(\infty\), \(-2b^{-1/2}\) approaches 0. So, the limit is \(2\).

Step 6: Confirm the Area

Since the limit converges to a finite number, the area between the curve \(y=\frac{1}{x^{3/2}}\) and the \(x\)-axis from \(x=1\) to \(\infty\) is 2.

Key Concepts

Area Under a Curve

Finding the area under a curve from one point to another is a fundamental concept in calculus. In simple terms, this involves calculating the sum of all the tiny squares between the curve and the x-axis. When dealing with functions like \( y = \frac{1}{x^{3/2}} \), which stretch infinitely, calculating this area requires special techniques. This area is represented by the definite integral of the function.

• The curve \( y = \frac{1}{x^{3/2}} \) decreases as \( x \) increases.
• We're interested in the area from \( x = 1 \) to \( x = \infty \).
• This necessitates the use of an improper integral because it extends to infinity.

In practical terms, we set up the problem using the integral \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \) to find this area. As we calculate it, we essentially find how the graph compares to the x-axis, figuring out how much space lies underneath.

Antiderivative

The concept of an antiderivative, or indefinite integral, is about finding a function whose derivative is the function you're analyzing. For our curve \( y = \frac{1}{x^{3/2}} \), an antiderivative is a function \( F(x) \) such that \( F'(x) = \frac{1}{x^{3/2}} \).

• Convert \( \frac{1}{x^{3/2}} \) to \( x^{-3/2} \).
• The antiderivative of \( x^{-3/2} \) is \( \frac{x^{-1/2}}{-1/2} = -2x^{-1/2} \).
• This means that when you differentiate \( -2x^{-1/2} \), you get back to \( \frac{1}{x^{3/2}} \).

By understanding this concept, we can evaluate the integral, thus finding the total area from one point to another under the curve. The process of taking antiderivatives is crucial for solving integrals, especially the improper ones which occur in calculus quite often.

Limit of Integration

When dealing with integrals that stretch to infinity, the limit of integration plays a vital role. Improper integrals, like \( \int_{1}^{\infty} \frac{1}{x^{3/2}} \, dx \), are managed through limits to handle the infinity aspect.

• Replace \( \infty \) with a variable, like \( b \), to keep calculations finite.
• Evaluate the integral from \( x = 1 \) to \( x = b \) first: \( \int_{1}^{b} \frac{1}{x^{3/2}} \, dx \).
• Calculate \( \lim_{{b \to \infty}} \left( \int_{1}^{b} \right) \) to find the actual answer.

In our example, after solving the integral, we find that the expression \( -2b^{-1/2} + 2 \) simplifies as \( b \) approaches infinity. The term \( -2b^{-1/2} \) fades away to zero, leaving the area to be \( 2 \). This final step confirms the infinite stretch has been accurately accounted for by the limit of integration.