Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int_{0}^{4} \frac{5}{\sqrt{2 t+1}} d t\)

Step-by-Step Solution

Answer

The integral evaluates to 10.

Step 1: Identify the integral to be solved

We are asked to find the integral of \( \int_{0}^{4} \frac{5}{\sqrt{2t+1}} \, dt \). The definite integral is calculated from the lower limit 0 to the upper limit 4.

Step 2: Recognize the standard integration formula

Notice that the integrand \( \frac{5}{\sqrt{2t+1}} \) can be approached using the formula for \( \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C \). Here, substitution is the key.

Step 3: Perform substitution

Let \( u = 2t + 1 \). Then, \( \frac{du}{dt} = 2 \) or \( dt = \frac{du}{2} \). This substitution helps transform the integral into a standard form.

Step 4: Change the limits of integration

When \( t = 0 \), \( u = 2(0) + 1 = 1 \). When \( t = 4 \), \( u = 2(4) + 1 = 9 \). So, the limits of integration change from 0 to 4 to 1 to 9.

Step 5: Rewrite the integral in terms of \( u \)

Substitute into the integral: \( \int_{1}^{9} \frac{5}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{5}{2} \int_{1}^{9} u^{-1/2} \, du \).

Step 6: Integrate the function

Using the integration formula \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = -1/2 \), calculate the integral: \( \frac{5}{2} \left[ \frac{u^{1/2}}{1/2} \right]_{1}^{9} = 5 \left[ \sqrt{u} \right]_{1}^{9} \).

Step 7: Evaluate the definite integral

Evaluate the result: \( 5 \left[ \sqrt{9} - \sqrt{1} \right] = 5 \left[ 3 - 1 \right] = 5 \times 2 = 10 \).

Key Concepts

Definite Integrals

Definite integrals provide a means to calculate the accumulation of quantities, such as areas under curves, between specified limits. In the expression \( \int_{a}^{b} f(x) \, dx \), the lower limit \( a \) and the upper limit \( b \) define the interval over which the integral is evaluated. Unlike indefinite integrals, which include a constant of integration \( C \), definite integrals result in a specific numerical value. This value represents the total change or area, considering the limits.

• The process involves taking a function, \( f(x) \), and evaluating its accumulation from \( a \) to \( b \).
• Important: The limits of integration determine where this area ends and begins.

When evaluating definite integrals, transformations such as substitutions often change the boundaries, but the integral value accounts for these adjustments.

U-Substitution Method

The u-substitution method is a technique used to simplify the process of integration, making complex integrals more manageable. It involves replacing part of the original integral with a single variable, \( u \), leading to a simpler expression. Here's a breakdown of the method:

• Identify the substitution: Choose a substitution \( u \), that simplifies the integrand. For example, in \( \int \frac{5}{\sqrt{2t+1}} \, dt \), setting \( u = 2t + 1 \) helps reduce the expression.
• Differentiate: Compute \( \frac{du}{dt} \) and rearrange to solve for \( dt \).
• Transform the integral: Substitute \( u \) and \( dt \) into the original integral.

After transforming, integrate with respect to \( u \), and substitute back the original variable to find your solution. This method simplifies integrals into standard forms, enabling easier and often faster solutions.

Integration Formulas

Integration relies on various formulas that serve as shortcuts for finding anti-derivatives (the reverse process of differentiation). These formulas are crucial in calculus as they provide direct solutions for common types of integrands.

• Basic Integral Formulas: Examples include \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \), \( \int e^x \, dx = e^x + C \), etc.
• Special Integrals: The formula used in our problem, \( \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C \), shows how certain integrals map directly to simpler functions.

Using these formulas, we simplify the calculation by replacing complex expressions or forms with known results. This streamlines not just the computation, but also increases our understanding of the integral's behavior.

Limits of Integration

Limits of integration define the scope of analysis for a definite integral. These are the values at which you start and stop calculating the area or accumulation. In definite integrals, once we perform a substitution like in u-substitution, these limits must change according to the new variable.

• Determine New Limits: When we substitute \( u = 2t + 1 \), the limits change from \( t \)-space to \( u \)-space. So, original \( t=0 \) becomes \( u=1 \), and \( t=4 \) becomes \( u=9 \).
• Evaluate with New Limits: Solve the integral using these transformed limits.

The key importance is calculating these correctly, as they ensure that the integral calculates the correct total area or value over the new transformed interval.