Problem 4 - Thomas Calculus

Question

The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{0}^{4} \frac{d x}{\sqrt{4-x}}$$

Step-by-Step Solution

Answer

The value of the integral is 4.

Step 1: Analyze the Integral

The given integral is \( \int_{0}^{4} \frac{dx}{\sqrt{4-x}} \). This is a definite integral, requiring evaluation over the interval from 0 to 4. The integrand \( \frac{1}{\sqrt{4-x}} \) suggests a substitution method, given the square root expression.

Step 2: Apply Substitution

To simplify the integrand \( \frac{1}{\sqrt{4-x}} \), use the substitution \( u = 4-x \). Then, \( du = -dx \), or \( dx = -du \). When \( x = 0 \), \( u = 4 \), and when \( x = 4 \), \( u = 0 \). Substitute these into the integral to get: \[-\int_{4}^{0} \frac{du}{\sqrt{u}}.\] Reversing the limits changes the sign: \[\int_{0}^{4} \frac{du}{\sqrt{u}}.\]

Step 3: Simplify the Integral

The integral \( \int_{0}^{4} \frac{du}{\sqrt{u}} \) can be rewritten as \( \int_{0}^{4} u^{-1/2} \, du \). This form is ready to be integrated using the power rule for integration.

Step 4: Integrate using the Power Rule

Apply the power rule \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for \( n eq -1 \) on \( u^{-1/2} \). Thus, \( n = -1/2 \), and the integral becomes: \[ \int u^{-1/2} \, du = \left[ 2u^{1/2} \right]_{0}^{4}.\]

Step 5: Evaluate the Integral at the Bounds

Substitute the limits into the result from the previous step: \[ 2\sqrt{4} - 2\sqrt{0} = 2 \times 2 - 2 \times 0 = 4. \] Therefore, the value of the definite integral is 4.

Key Concepts

Substitution Method

The substitution method is a very handy technique for evaluating integrals, especially when dealing with complex expressions like square roots. The core idea is to simplify the integrand (the function you're integrating) by making a substitution that makes the integral easier to evaluate.

• First, choose a substitution that replaces a part of the integrand. For the integral \( \int \frac{dx}{\sqrt{4-x}} \), the substitution \( u = 4-x \) simplifies the square root expression.
• The corresponding derivative, \( du = -dx \), allows us to replace \( dx \) with \(-du\).
• Adjust the limits of integration. The original limits for \( x \) were 0 and 4, which change to 4 and 0 for \( u \) after the substitution.

Flipping the limits' order can change the integral's sign, which is corrected by reversing the limits, as seen when the integral becomes \( \int_{0}^{4} \frac{du}{\sqrt{u}} \). This process simplifies the original problem into a more manageable form.

Power Rule for Integration

The power rule for integration is one of the fundamental tools in calculus, used to integrate functions of the form \( u^n \). It states that if \( n eq -1 \), the integral of \( u^n \) with respect to \( u \) is \( \frac{u^{n+1}}{n+1} + C \), where \( C \) represents the constant of integration.

• Applying this rule to our integral \( \int u^{-1/2} \, du \), we identify \( n = -1/2 \).
• Using the formula, the integral becomes \( \frac{u^{1/2}}{1/2} + C \), simplifying to \( 2u^{1/2} + C \).

For definite integrals like this one, the constant of integration \( C \) is not needed. You simply evaluate the antiderivative at the upper and lower bounds of the integral. This makes the power rule a quick and effective method for finding integrals when the function fits the pattern.

Integrals with Square Roots

Handling integrals with square roots can be challenging, but by transforming them into simpler expressions, they become more approachable. The presence of the square root \( \sqrt{4-x} \) in the original problem suggests a substitution like \( u = 4-x \), which turns the square root into \( \sqrt{u} \).This transformation:

• Replaces the expression under the square root, \( \sqrt{4-x} \), with a simpler variable, \( \sqrt{u} \).
• The resulting integral \( \int u^{-1/2} \, du \) is much easier to tackle using familiar techniques like the power rule.

By reducing the complexity of the expression inside the square root, substitution allows easier integration and clearer understanding of the integral's behavior over its limits. This makes managing integrals with square roots not just feasible but straightforward with practice.