Problem 54 - Calculus and its applications

Question

Another form of improper integral occurs when a vertical asymptote appears at one of the bounds of integration. For example, to find the area under \(f(x)=\frac{1}{\sqrt{x}}\) over \([0,4],\) we note that \(f\) is not defined at \(x=0\) and that there is a vertical asymptote as \(x\) approaches 0 from the right. In such a case, we integrate over \([a, 4]\) and find the limit as a approaches 0 from the right Use this technique. $$ \text { Find } \int_{4}^{5} \frac{3}{\sqrt[3]{x-5}} d x \text { . } $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{9}{2} \).

Step 1: Understand the Problem

We need to evaluate the integral \( \int_{4}^{5} \frac{3}{\sqrt[3]{x-5}} \, dx \). The function \( f(x) = \frac{3}{\sqrt[3]{x-5}} \) presents a vertical asymptote at \( x=5 \) because the denominator approaches zero as \( x \to 5 \) from the left.

Step 2: Set Up the Improper Integral

To handle the vertical asymptote at \( x=5 \), introduce a variable \( a \) approaching 5 from the left and rewrite the integral from 4 to \( a \): \[ \int_{4}^{5} \frac{3}{\sqrt[3]{x-5}} \, dx = \lim_{a \to 5^-} \int_{4}^{a} \frac{3}{\sqrt[3]{x-5}} \, dx \]

Step 3: Find the Antiderivative

Consider \( u = x - 5 \), then \( du = dx \). The function becomes \( \int \frac{3}{u^{1/3}} \, du \). The antiderivative of \( u^{-1/3} \) is \( \frac{3}{2} u^{2/3} \). Thus, \[ \int \frac{3}{u^{1/3}} \, du = 3 \cdot \frac{3}{2} u^{2/3} = \frac{9}{2} u^{2/3} + C \]

Step 4: Substitute Back and Simplify

Substitute \( u = x - 5 \) back into the antiderivative to get:\[ \int \frac{3}{\sqrt[3]{x-5}} \, dx = \frac{9}{2} (x-5)^{2/3} + C \]

Step 5: Evaluate the Limits

Evaluate the definite integral at \( 4 \) and \( a \): \[ \lim_{a \to 5^-} \left( \frac{9}{2} (a-5)^{2/3} - \frac{9}{2} (4-5)^{2/3} \right) \]Substitute \( b = a - 5 \), as \( b \to 0^- \), \( \frac{9}{2} b^{2/3} \to 0 \), thus:\[ \lim_{a \to 5^-} \left( \frac{9}{2} (a-5)^{2/3} \right) = 0 \].Substitute -1 for \( (4-5) \) and simplify:\[ -\frac{9}{2} \cdot (-1)^{2/3} = -\frac{9}{2} \]

Step 6: Final Calculation

With the limit evaluated, the original expression becomes:\[ 0 - \left(-\frac{9}{2}\right) = \frac{9}{2} \]

Key Concepts

Vertical asymptote

When we talk about vertical asymptotes, you can imagine it as the line where the function shoots off to infinity, either positively or negatively, as it approaches this line. - **Real-life Function**: In the given problem with the function \( f(x) = \frac{3}{\sqrt[3]{x-5}} \), we see a vertical asymptote at the point \( x = 5 \). Why does this happen? As \( x \) approaches 5 from the left (which we denote as \( x \to 5^- \)), the denominator \( \sqrt[3]{x-5} \) gets closer and closer to zero. - **Fraction Effect**: Whenever you divide by a very small number, the fraction's value grows larger and larger (approaches infinity). That's what creates the asymptote.Understanding vertical asymptotes is crucial in calculus, especially when dealing with improper integrals. The function's behavior around vertical asymptotes dictates how we approach the integral limits, keeping our calculations accurate and meaningful.

Definite integral

A definite integral helps us find the area under a curve between two bounds. In calculus, this is like measuring the exact covered area by the graph of a function from one point to another on the x-axis. - **With Asymptotes**: When there's a vertical asymptote involved, the integral is no longer straightforward. This is known as an improper integral because the function isn't defined at one of the bounds or within the interval—for instance, our bound at \( x = 5 \).- **Evaluation**: One crucial step is finding the limit of this area as it approaches the asymptote. In practical terms, for our integral \( \int_{4}^{5} \frac{3}{\sqrt[3]{x-5}} \, dx \), we split it into a limit problem, making the integration process smoother by avoiding direct integration at the undefined point.By understanding and rewriting the integral in this way, we work around the vertical asymptote to accurately solve for the total area or value of the definite integral.

Limit of integration

The limit of integration is a key concept when solving definite integrals, especially when encountering vertical asymptotes. - **Approaching the Asymptote**: Instead of directly integrating to the problematic point where the function is undefined (in our case \( x = 5 \)), we let a variable approach this point, say \( a \to 5^- \). This creates a new limit expression: \[ \lim_{a \to 5^-} \int_{4}^{a} \frac{3}{\sqrt[3]{x-5}} \, dx \]- **Why Take Limits**: The limit helps us approach the true behavior of the area under the curve as closely as possible, letting us "sneak up" on the vertical asymptote without actually hitting it, which makes the math work out nicely as the area approaches a finite value.- **From Setup to Solution**: By setting up our integral with a limit, we transition from a problematic improper scenario to a solvable limit problem, allowing us to find the area reliably and correctly calculate the actual numerical value.