Problem 46 - University Calculus: Early Transcendentals

Question

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises \(1-46\). $$\int_{0}^{3} \frac{y d y}{\sqrt{5 y+1}}$$

Step-by-Step Solution

Answer

The integral evaluates to 7.2.

Step 1: Identify Substitution Variable

The original integral is \( \int_{0}^{3} \frac{y \, dy}{\sqrt{5y+1}} \). Here, we choose \( u = 5y + 1 \) as the substitution variable to simplify the integrand.

Step 2: Calculate Derivative

Differentiate the substitution variable \( u = 5y + 1 \) with respect to \( y \), giving \( \frac{du}{dy} = 5 \). Thus, \( dy = \frac{du}{5} \).

Step 3: Change Limits of Integration

As the substitution changes the variable from \( y \) to \( u \), the limits also need to be converted. When \( y=0 \), \( u = 5(0) + 1 = 1 \). When \( y=3 \), \( u = 5(3) + 1 = 16 \). So, the limits of integration are from 1 to 16.

Step 4: Substitute and Simplify Integral

Substitute \( u \) and \( dy \) into the integral: \( \int_{1}^{16} \frac{y}{\sqrt{u}} \cdot \frac{du}{5} \). To express \( y \) in terms of \( u \), rearrange the substitution formula: \( y = \frac{u - 1}{5} \). Substitute to get \( \frac{1}{5} \int_{1}^{16} \frac{u - 1}{\sqrt{u}} \, du \).

Step 5: Separate and Simplify the Integral

Separate the integral into two: \( \frac{1}{5} \left( \int_{1}^{16} u^{1/2} \, du - \int_{1}^{16} u^{-1/2} \, du \right) \).

Step 6: Evaluate the Integrals

For \( \int u^{1/2} \, du \), use the power rule: the integral is \( \frac{2}{3} u^{3/2} \) from 1 to 16. For \( \int u^{-1/2} \, du \), the integral is \( 2u^{1/2} \) from 1 to 16.

Step 7: Evaluate at the Boundaries

Calculate \( \frac{1}{5} \left[ \left( \frac{2}{3}(16)^{3/2} - \frac{2}{3}(1)^{3/2} \right) - \left( 2(16)^{1/2} - 2(1)^{1/2} \right) \right] \).

Step 8: Final Calculation

Compute: \( \frac{2}{3}(16^{3/2}) = \frac{2}{3}(64) = \frac{128}{3} \) and \( \frac{2}{3}(1^{3/2}) = \frac{2}{3} \). Also, \( 2(16^{1/2}) = 2(4) = 8 \) and \( 2(1^{1/2}) = 2 \). So, the expression becomes \( \frac{1}{5} \left[ \frac{128}{3} - \frac{2}{3} - 8 + 2 \right] = \frac{1}{5} \left[ \frac{126}{3} - 6 \right] = \frac{1}{5} \left[42 - 6\right] = \frac{1}{5}(36) = 7.2 \).

Key Concepts

Substitution Method

The substitution method is a fundamental technique used in integral calculus to simplify the process of finding integrals. The basic idea is to make a substitution that transforms a difficult integral into an easier one.
By changing variables, we can often match the structure of the integral to a basic formula or rule that is easier to integrate. This is particularly useful when dealing with complex functions and multiple variables. In our original exercise, we use the substitution method to simplify the integral \( \int_0^3 \frac{y \, dy}{\sqrt{5y+1}} \).
To apply the substitution, we choose a new variable \( u \) such that \( u = 5y + 1 \). This change of variables transforms the original expression into an integral in terms of \( u \) rather than \( y \).
Once the substitution is complete, the integrand becomes something much easier to integrate.Substitute your chosen variable back into the function, and remember to adjust your limits of integration as your variable changes. This forms the basis of the power and practicality of substitution in calculus.

Integration by Substitution

Integration by substitution, also known as "u-substitution," is essentially the reverse of the chain rule in differentiation. The primary goal is to simplify the integrand to an easier form for integration.
This method relies heavily on choosing the correct substitution that efficiently simplifies the integration process.Let’s delve into the process of integration by substitution using our example:

• First, identify the part of the integrand that can be replaced with a simpler expression. We selected \( u = 5y + 1 \).
• Next, find the differential of \( u \) with respect to the original variable. In this case, the derivative \( du = 5dy \) means \( dy = \frac{du}{5} \).
• Substitute \( u \) and \( dy \) into the integral: our expression becomes \( \int_1^{16} \frac{u - 1}{5 \sqrt{u}} \, du \).

The whole essence is reducing the integral into a form you're familiar with, usually leading to a straightforward application of the power rule or other basic integral formulas. After integrating, remember to substitute back the original variable unless dealing with definite integrals, where adjusted limits address this.

Definite Integrals

Definite integrals are very useful in calculus, giving the area under a curve between two specific points. They are denoted by the notation \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the boundaries of integration. Unlike indefinite integrals, which represent a family of functions, definite integrals result in a numerical value.In the context of the example, because we are dealing with definite integrals, the limits of integration change along with the substitution. Originally, the limits were from \( y = 0 \) to \( y = 3 \). However, after substituting \( u = 5y + 1 \), we had to adjust these limits as follows:

• For \( y = 0 \), substitute back to get \( u = 1 \).
• For \( y = 3 \), it becomes \( u = 16 \).

With these new limits, the integration is performed in terms of \( u \) from 1 to 16. This ensures that the final result corresponds correctly to the original integral's interval.Definite integrals provide a specific value that represents the accumulated "sum" over the interval, reflecting concepts like total change, area, and accumulated distance.