Problem 17 - Calculus for Biology and Medicine

Question

In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \sqrt{x+2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2}{3}(x+2)^{3/2} + C \).

Step 1: Identify the Substitution

Identify the expression inside the square root to use for substitution. Let \( u = x + 2 \).

Step 2: Calculate \( du \)

Differentiate \( u \) with respect to \( x \) to find \( du \). Since \( u = x+2 \), we have \( \frac{du}{dx} = 1 \), which gives \( du = dx \).

Step 3: Substitute in the Integral

Replace \( x + 2 \) with \( u \) and \( dx \) with \( du \). The integral becomes \( \int \sqrt{u} \, du \).

Step 4: Evaluate the New Integral

Find the integral of \( \sqrt{u} \). Recall that \( \sqrt{u} = u^{1/2} \), so integrate \( u^{1/2} \) using the power rule: \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C \).

Step 5: Simplify the Expression

Simplify \( \frac{u^{3/2}}{3/2} \) to \( \frac{2}{3}u^{3/2} + C \).

Step 6: Substitute Back to the Original Variable

Replace \( u \) with \( x+2 \). The solution is \( \frac{2}{3}(x+2)^{3/2} + C \).

Key Concepts

Indefinite Integrals

Indefinite integrals are a foundational concept in calculus. They represent the collection of all antiderivatives of a given function. An indefinite integral does not have specified limits, which means it can represent a broad range of functions.The notation for an indefinite integral is:\[\int f(x) \, dx\]Instead of being just a number, an indefinite integral gives a whole family of functions. Each function differs by a constant, denoted as \( C \), often referred to as the "constant of integration."

• The indefinite integral of a function \( f(x) \) is essentially the reverse process of differentiation.
• The constant \( C \) ensures that the function can shift up or down on a graph without affecting its derivative.

When solving problems involving indefinite integrals, we often use integration techniques such as substitution to simplify the process and find the antiderivative efficiently.

Substitution Method

The substitution method is a powerful technique used to solve integrals that are not readily integrable in their current form. This method involves changing the variable of integration to simplify the integral.To effectively use substitution, follow these steps:1. **Identify the substitution**: Choose a new variable (usually \( u \)) that represents a more complicated part of the integral. In the exercise, the expression inside the square root \( x+2 \) is chosen as \( u \).2. **Differentiate to find \( du \)**: Differentiate the chosen \( u \) with respect to \( x \) to obtain \( du \). This ensures that you can properly replace \( dx \) in the integral. By letting \( u = x+2 \), differentiating gives \( du = dx \).3. **Substitute and transform**: Substitute \( u \) and \( du \) into the integral to transform it into a simpler form. The original integral \( \int \sqrt{x+2} \, dx \) becomes \( \int \sqrt{u} \, du \) after substitution.Remember that after integrating with respect to \( u \), substitute back in terms of the original variable \( x \) to express the final solution accurately.

Power Rule in Integration

The power rule in integration is akin to the power rule in differentiation, but used in reverse. It is pivotal for integrating functions of the form \( x^n \), where \( n eq -1 \).The rule states that:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Applying this rule effectively requires paying attention to the new power after integration. Here's how we used it in the step-by-step solution:- Since \( \sqrt{u} = u^{1/2} \), applying the power rule to integrate \( u^{1/2} \, du \) involves increasing the exponent by 1 (so \( 1/2 + 1 = 3/2 \)) and dividing by this new exponent.- This gave us \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C \).This process highlights why the power rule is essential when dealing with polynomial and radical expressions in integration. Understanding and mastering it allows for efficient solving of integrals without trial and error.