Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{5}^{8} \sqrt{3 x+1} d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \\(\frac{122}{9}\\).

Step 1: Identify the Substitution

We look at the integrand \(\sqrt{3x+1}\) and decide to substitute \(u = 3x + 1\), which is what's inside the square root. This substitution will simplify the integral.

Step 2: Calculate the Derivative/ Differential

Differentiate the substitution equation \(u = 3x + 1\) with respect to \(x\), which gives \(\frac{du}{dx} = 3\). Thus, \(du = 3 dx\) or \(dx = \frac{1}{3} du\).

Step 3: Change the Limits of Integration

Change the limits of integration from \(x\) to \(u\). When \(x=5\), \(u = 3(5) + 1 = 16\); when \(x=8\), \(u = 3(8) + 1 = 25\). So the new integral will be with limits from \(u = 16\) to \(u = 25\).

Step 4: Perform the Substitution

Substitute for \(u\) and \(dx\) in the integral: \(\int_{5}^{8} \sqrt{3x+1} \, dx = \int_{16}^{25} \sqrt{u} \left(\frac{1}{3}\right) \, du\).

Step 5: Simplify the Integral

Simplify the integral to \(\frac{1}{3} \int_{16}^{25} u^{1/2} \, du\).

Step 6: Integrate

Integrate \(u^{1/2}\) to get \(\frac{2}{3} u^{3/2}\). So, we have: \(\frac{1}{3} \left[ \frac{2}{3} u^{3/2} \right]_{16}^{25}\).

Step 7: Evaluate the Definite Integral

Evaluate the expression \(\frac{1}{3} \times \frac{2}{3} \left[25^{3/2} - 16^{3/2}\right]\) which becomes \(\frac{2}{9} \left[125 - 64\right] = \frac{2}{9}(61) = \frac{122}{9}\).

Key Concepts

Definite Integrals

Definite integrals represent the area under the curve of a function on a specific interval. Unlike their indefinite counterparts, definite integrals have both upper and lower limits. These limits create a boundary for the integration and are critical in determining the exact area. In our exercise, the definite integral is represented as \( \int_{5}^{8} \sqrt{3x+1} \, dx \). Here, we are looking at the function \( \sqrt{3x+1} \) between the x-values of 5 and 8.
This means we are calculating the exact accumulated change from the point where \( x = 5 \) to \( x = 8 \). Unlike finding a general form of an antiderivative, definite integrals produce a numerical value, representing the total area under the curve within the specified limits. Remember, the key components in a definite integral include:

• The integrand, which is the function being integrated (e.g., \( \sqrt{3x+1} \)).
• The limits of integration, which define the interval (e.g., from 5 to 8).
• The differential (e.g., \( dx \)), which indicates the variable with respect to which integration is performed.

Mastering definite integrals is vital, as they form the backbone of topics involving area, volume, and other physical applications in calculus.

Integration Techniques

Integration techniques are essential tools for evaluating more complex integrals. One of the most frequently used techniques is substitution, especially when addressing integrals that include composite functions. These techniques help simplify the integrand, making it easier to integrate.
In calculus, several integration techniques are employed, such as:

• Substitution: This is utilized when the derivative of a function within the integral is present elsewhere in the integrand, as demonstrated in our exercise.
• Integration by parts: Useful for products of functions, applying the formula \( \int u \, dv = uv - \int v \, du \).
• Partial fraction decomposition: Applied primarily to rational functions, breaking them into simpler fractions.
• Trigonometric identities: Often used for integrands involving trigonometric functions.

These methods stand essential when typical antiderivation processes prove challenging. Our focus in this exercise is on substitution, which typically involves identifying a part of the integrand that can be replaced, simplifying the process.

Substitution in Calculus

Substitution in calculus involves altering the integral's variable and replacing it with a new expression or variable to make the integral easier to evaluate. This method is especially powerful for integrals that involve nested functions or a composition of functions that would be difficult to integrate directly.
The process of substitution generally follows these steps:

• Choose a substitution: Identify a part of the integral that can be set as \( u \). For example, in our problem, \( u = 3x + 1 \) was selected since it's inside the square root.
• Differentiate: Find \( \frac{du}{dx} \) and rearrange to solve for \( dx \), giving \( dx = \frac{1}{3} du \) in our exercise.
• Transform the limits: The original limits in terms of \( x \) must be converted to \( u \). When \( x = 5 \), then \( u = 16 \); and when \( x = 8 \), then \( u = 25 \).
• Rewrite the integral: Replace all parts in terms of \( u \) and \( du \), leading to a simpler integral to solve.

Substitution allows a tricky integral to be transformed into something more manageable, lending itself to easier integration and evaluation.

Limits of Integration

Limits of integration define the interval over which the definite integral is evaluated. They specify where the integration starts and ends, setting a finite boundary. These limits are vital in substituting definite integrals because, upon changing variables, limits must also be recalculated to fit the new variable.
In the given exercise, our original limits were \( x = 5 \) to \( x = 8 \). After substitution, these transformed into \( u = 16 \) to \( u = 25 \). Calculating limits is crucial to ensure the evaluation reflects the area over the same interval as initially presented.

• Transforming limits ensures coherence with the new substitution variable. This step ensures we compute the integral over the same respective segment on the variable's graph.
• Always double-check limits after substitution to avoid errors in the integral's evaluation.

The correct application of transformation rules for limits ensures the definite integral’s outcome accurately captures the bounded area under the curve for the substituted function.