Question

Evaluate the improper integrals. Each of these integrals has an infinite discontinuity either at an endpoint or at an interior point of the interval.\(\int_{0}^{9} \frac{d x}{\sqrt{9-x}}\)

Short answer

The value of the improper integral is 6.

Step-by-step solution

Identify the Type of Improper Integral

The integral \( \int_{0}^{9} \frac{d x}{\sqrt{9-x}} \) is an improper integral because there is an infinite discontinuity at the upper limit of integration (as \( x \rightarrow 9 \), the denominator of the integrand goes to zero, creating a division by zero issue).

Set Up the Limit Definition

To handle the discontinuity at \( x = 9 \), we need to rewrite the integral using limits. This means we will evaluate the limit as \( b \) approaches 9 from the left:\[ \lim_{{b \to 9^-}} \int_{0}^{b} \frac{dx}{\sqrt{9-x}}. \]

Perform the Integration

The antiderivative of \( \frac{1}{\sqrt{9-x}} \) with respect to \( x \) is \( -2\sqrt{9-x} \) (this can be found using a substitution method or by recognizing the standard integral form).Thus, the integral from \( 0 \) to \( b \) becomes:\[ \int_{0}^{b} \frac{dx}{\sqrt{9-x}} = \left[-2\sqrt{9-x}\right]_0^b. \]

Evaluate the Definite Integral

Substituting the limits of integration, we have:\[ -2\sqrt{9-b} + 2\sqrt{9-0} = -2\sqrt{9-b} + 6. \]

Calculate the Limit

We now take the limit as \( b \) approaches 9 from the left:\[ \lim_{{b \to 9^-}} \left( -2\sqrt{9-b} + 6 \right). \]As \( b \to 9 \), \( \sqrt{9-b} \) tends to zero, making the expression tend to:\[ -2 \cdot 0 + 6 = 6. \]

Key concepts

Infinite Discontinuity

In mathematics, infinite discontinuity occurs when a function approaches an infinite value at a specific point or over an interval. This typically happens when the denominator of a function approaches zero, which causes the function to "blow up" to infinity. In the context of improper integrals, such as the one in our exercise, an infinite discontinuity is present at the endpoint where the integrand becomes undefined.
In this specific case, at the upper limit (9), the denominator \( \sqrt{9-x} \) approaches zero as \( x \rightarrow 9 \). This results in a scenario where the function has a division by zero, creating the discontinuity at that point.
To solve an integral with such features, we use techniques like the limit definition to manage the discontinuity and properly evaluate the integral.

Limit Definition

The limit definition is a crucial tool in handling improper integrals that exhibit infinite discontinuities. Instead of directly integrating over an interval with potential infinity conditions, we redefine the integral by introducing a limit. This allows us to carefully examine the behavior of the function as it approaches the problematic point.
For the current scenario, where the integrand \( \frac{1}{\sqrt{9-x}} \) becomes problematic at \( x = 9 \), we express the integral using a limit. Specifically, we evaluate the limit of the integral as the variable \( b \) nears 9 from the left:

• \( \lim_{{b \to 9^-}} \int_{0}^{b} \frac{dx}{\sqrt{9-x}} \)

This alters the way we calculate the integral, focusing on the behavior as \( x \) approaches 9, ensuring accuracy in the evaluation. This technique is fundamental in calculus, especially when dealing with boundaries that include infinite or undefined behavior.

Antiderivative

An antiderivative is a function whose derivative yields the original function that was being integrated. Obtaining the antiderivative is a key step in computing definite integrals. In the given problem, we needed to find the antiderivative of \( \frac{1}{\sqrt{9-x}} \). This can often be done through pattern recognition or methods like substitution, recognizing it as a standard form of integral.
The antiderivative for this expression is \( -2\sqrt{9-x} \). Once we have this antiderivative, we can apply the fundamental theorem of calculus to move forward with computing the definite integral.
Knowing how to derive antiderivatives is crucial for solving not only improper integrals but all definite integrals in calculus.

Substitution Method

The substitution method is a technique used to simplify the integration process. It involves changing the variable of integration to reduce the complexity of the integral.
In the case of the integral \( \frac{1}{\sqrt{9-x}} \), a substitution can help clarify the operation. For instance, if we let \( u = 9-x \), then \( du = -dx \), which transforms the way we deal with the integral.

• This simplifies our problem into standard forms which are easier to integrate.
• Such substitutions often enable us to resolve complex integrands into more manageable forms by transforming into known integral forms.

Successfully using substitutions allows us to find antiderivatives more intuitively, helping us solve problems like the improper integral at hand more efficiently.