Question

Evaluate each improper integral or show that it diverges. \(\int_{4}^{\infty} \frac{d x}{(\pi-x)^{2 / 3}}\)

Short answer

The integral diverges.

Step-by-step solution

Identify the type of improper integral

The integral \( \int_{4}^{\infty} \frac{d x}{(\pi-x)^{2 / 3}} \) is an improper integral because it has an infinite upper limit, approaching infinity as \( x \) tends to infinity.

Set up the limit to evaluate the integral

To evaluate the integral, we transform it into a limit: \[ \lim_{b \to \infty} \int_{4}^{b} \frac{d x}{(\pi-x)^{2 / 3}}. \] We will first focus on solving \( \int_{4}^{b} \frac{d x}{(\pi - x)^{2 / 3}} \) and then take the limit as \( b \to \infty \).

Make a substitution to simplify the integral

Let \( u = \pi - x \), then \( du = -dx \), or \( dx = -du \). The limits of integration change from \( x = 4 \) to \( u = \pi - 4 \) and \( x = b \) to \( u = \pi - b \). The integral becomes:\[ -\int_{\pi-b}^{\pi-4} \frac{du}{u^{2/3}}. \]

Integrate using the power rule

The antiderivative of \( u^{-2/3} \) is \( u^{1/3} \cdot \frac{3}{1} = 3u^{1/3} \). So, the integral is:\[ -3 \left[ u^{1/3} \right]_{\pi-b}^{\pi-4}. \]

Evaluate the definite integral

Substitute the limits back into the antiderivative:\[ -3\left((\pi-4)^{1/3} - (\pi-b)^{1/3}\right). \] This simplifies to:\[ 3\left((\pi-b)^{1/3} - (\pi-4)^{1/3}\right). \]

Take the limit as b approaches infinity

Consider the limit:\[ \lim_{b \to \infty} 3\left((\pi-b)^{1/3} - (\pi-4)^{1/3}\right). \] As \( b \to \infty \), \( \pi-b \to -\infty \). The term \( (\pi-b)^{1/3} \to -\infty^{1/3} = -\infty \). Therefore, the expression diverges to negative infinity, meaning the entire integral does not converge.

Key concepts

Limits in Calculus

When working with integrals, especially improper integrals, understanding limits in calculus is vital. Improper integrals often involve integrands with infinite limits or undefined behavior at certain points. These situations require us to use limits to evaluate the behavior of the function approaching these points.

In our exercise, we dealt with an integral having an upper limit of infinity. To handle this, we transformed the integral into:

• \[ \lim_{b \to \infty} \int_{4}^{b} \frac{d x}{(\pi-x)^{2/3}} \]

This manipulation allows us to focus on the finite section between 4 and a variable value 'b', which will ultimately tend toward infinity. By evaluating the definite integral, we finally examine its behavior as the upper limit ( 'b') approaches infinity.

This is where the concept of limits shines. Not only do they help us evaluate specific points, but they also help us understand the behavior of functions as variables approach particular values. In this case, as \( b \to \infty \).

Antiderivatives

Antiderivatives play a crucial role in solving integrals. Essentially, finding an antiderivative means identifying a function whose derivative is the given function. This is what helps in deducing the solution for definite integrals.

In our exercise, after transforming the original integral into terms of 'u', the focus was on finding the antiderivative of \( u^{-2/3} \).

By following the integration rules:

• The antiderivative of \( u^{-2/3} \) is \( u^{1/3} \cdot \frac{3}{1} = 3u^{1/3} \).
• This means, when we derive \( 3u^{1/3} \), we revert back to \( u^{-2/3} \).

Understanding and identifying antiderivatives is key to solving integrals and often serves as a foundation for almost any calculus problem involving integration.

Power Rule in Integration

The power rule in integration is a fundamental technique to find antiderivatives. This rule assists in integrating functions under a power form. If you have a function \( x^n \), using the power rule involves changing the exponent.

The power rule states:

• Integrate \( x^n \) to get \( \frac{x^{(n+1)}}{n+1} \)
• Applicable when \( n eq -1 \)

In terms of our problem, the integrand \( u^{-2/3} \) used the power rule. The integration leads us to \(3u^{1/3}\), which is a direct application of the power rule formula.

Knowing when and how to use the power rule is indispensable. It simplifies complicated expressions and enables calculus students to tackle various integration challenges efficiently. Understanding this rule is vital for gaining confidence and proficiency in integrating functions.