Problem 11 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{0}^{9} \frac{d x}{\sqrt{9-x}} $$

Step-by-Step Solution

Answer

The integral \( \displaystyle \int_{0}^{9} \frac{dx}{\sqrt{9-x}} \) is 6.

Step 1: Identify Why the Integral is Improper

The given integral is improper because the integrand \( \frac{1}{\sqrt{9-x}} \) becomes undefined when \( x = 9 \). This is due to the square root in the denominator, as \( \sqrt{9-9} = \sqrt{0} = 0 \), which makes the denominator zero, resulting in division by zero.

Step 2: Set Up the Limit for the Improper Integral

To handle the discontinuity at \( x = 9 \), we evaluate the integral as a limit: \[ \lim_{b \to 9^{-}} \int_{0}^{b} \frac{d x}{\sqrt{9-x}} \] This replaces the problematic upper bound with a variable \( b \) approaching 9 from the left.

Step 3: Perform the Substitution

Use the substitution \( u = 9 - x \), which leads to \( du = -dx \) or \( dx = -du \). Thus, the integral becomes: \[ -\lim_{b \to 9^{-}} \int_{9}^{9-b} \frac{d u}{\sqrt{u}} \] Changing the bounds accordingly: when \( x = 0, u = 9 \), and when \( x = b, u = 9 - b \).

Step 4: Integrate

The integral \( \int \frac{d u}{\sqrt{u}} \) can be rewritten as \( \int u^{-1/2} \, du \), which evaluates to \( 2u^{1/2} + C \). Thus, the specific integral is: \[ \lim_{b \to 9^{-}} -\left[ 2u^{1/2} \right]_{9}^{9-b} = \lim_{b \to 9^{-}} -\left( 2\sqrt{9-b} - 2\sqrt{9} \right) \]

Step 5: Evaluate the Limit

Now, calculate the limit:\[ \lim_{b \to 9^{-}} -\left( 2\sqrt{9-b} - 2\sqrt{9} \right) = -\left( 2\cdot0 - 6 \right) = 6 \] Therefore, the value of the integral is 6.

Key Concepts

Limit Evaluation

When dealing with improper integrals, like the integral \( \int_{0}^{9} \frac{d x}{\sqrt{9-x}} \), limit evaluation is a crucial tool. This type of integral is improper because of the discontinuity at the upper bound \( x = 9 \), where the function becomes undefined due to division by zero. To properly evaluate an improper integral like this, one approach is to express the problematic portion as a limit.
This involves replacing the upper limit with a variable, say \( b \), and then considering the limit as \( b \) approaches the problematic point, in this case 9, from the left.

• This process converts the integral into a limit expression: \[ \lim_{b \to 9^{-}} \int_{0}^{b} \frac{d x}{\sqrt{9-x}} \]
• The answer hinges on evaluating this limit, effectively sidestepping the undefined behavior at \( x = 9 \).

Understanding limit evaluation is essential in making use of the fundamental theorem of calculus in contexts where functions behave irregularly at certain boundaries.

Substitution Method

The substitution method is a powerful technique in calculus for simplifying and evaluating integrals. When faced with the integral \( \int_{0}^{9} \frac{d x}{\sqrt{9-x}} \), substitution helps by changing variables to make the integration easier.
For this particular problem, using substitution allows for simplifying the square root function in the denominator.

• We apply the substitution \( u = 9 - x \). Consequently, the differential \( du = -dx \), so \( dx = -du \).
• This translates the integral to: \[ -\lim_{b \to 9^{-}} \int_{9}^{9-b} \frac{d u}{\sqrt{u}} \]
• It's crucial to update the bounds according to \( u \'s\) relationship with \( x \): when \( x = 0, u = 9\) and when \( x = b, u = 9-b\).

This technique makes evaluating the resulting integral much simpler, allowing us to apply known methods to solve it.

Square Root Functions

Square root functions, like the \( \sqrt{9-x} \) seen in our integrand, can often present challenges, particularly when they appear in the denominator.
These functions have significant limitations, especially when the argument under the square root approaches zero, as this can lead to undefined behavior like division by zero.

• In this problem, the square root function evaluates to zero at \( x = 9 \), causing an improper integral and illustrating the need for careful handling.
• The integration becomes more manageable after substitution because it avoids direct confrontation with the zero in the denominator.

Understanding the behavior of square root functions is crucial, especially in improper integrals, as it informs the choice of techniques like substitution and underscores the role of limits to evaluate these expressions accurately.