Problem 54 - Calculus for Biology and Medicine

Question

In Problems 43-58, use substitution to evaluate each definite integral. $$ \int_{4}^{9} \frac{x}{x-3} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( 5 + 3 \ln 6 \).

Step 1: Choose a substitution

To evaluate the integral \( \int_{4}^{9} \frac{x}{x-3} \, dx \), start by choosing a substitution to simplify the denominator. Set \( u = x - 3 \).

Step 2: Differentiate the substitution

Differentiate the substitution \( u = x - 3 \) to find \( du \). This yields \( du = dx \).

Step 3: Change the limits of integration

Transform the limits of integration according to the substitution. When \( x = 4 \), \( u = 4 - 3 = 1 \). When \( x = 9 \), \( u = 9 - 3 = 6 \). Thus, the limits for \( u \) are from 1 to 6.

Step 4: Express the integrand in terms of u

Substitute \( u = x - 3 \) into the integrand, so \( x = u + 3 \). The integrand becomes \( \frac{x}{x-3} = \frac{u+3}{u} = 1 + \frac{3}{u} \).

Step 5: Rewrite the integral in terms of u

The integral now becomes \( \int_{1}^{6} \left( 1 + \frac{3}{u} \right) \, du \).

Step 6: Separate the integral

Split the integral into two separate integrals: \( \int_{1}^{6} 1 \, du \) and \( 3 \int_{1}^{6} \frac{1}{u} \, du \).

Step 7: Evaluate each integral

Evaluate the first integral: \( \int_{1}^{6} 1 \, du = [u]_{1}^{6} = 6 - 1 = 5 \). For the second integral, \( 3 \int_{1}^{6} \frac{1}{u} \, du = 3[\ln|u|]_{1}^{6} = 3[\ln 6 - \ln 1] = 3 \ln 6 \) since \( \ln 1 = 0 \).

Step 8: Sum the results

Add the results of the integrals: \( 5 + 3 \ln 6 \). Thus, the value of the original integral is \( 5 + 3 \ln 6 \).

Key Concepts

Substitution Method

The substitution method is a technique used in calculus to simplify the process of evaluating integrals. Here is how it works:

• First, identify a part of the integral that can be substituted with a new variable. This is usually a component that complicates the integration.
• The goal is to replace the variable and expression in the integrand with the new variable, making the integral easier to solve.

For example, given the integral \( \int \frac{x}{x-3} \, dx \), we can set \( u = x - 3 \). This simplifies the denominator, letting us express the integral in simpler terms of \( u \). The substitution \( u = x - 3 \) is chosen because it directly simplifies the expression. Differentiating this gives \( du = dx \), which helps in rewriting the integral.
Overall, the substitution method is your best friend when dealing with complicated integrals, as it turns them into more manageable problems.

Integral Limits

In definite integrals, adjusting the limits is an essential step when using substitution. Here's how you adjust the limits of integration:

• First, understand that the original limits (e.g., from \( x = 4 \) to \( x = 9 \)) must be converted to the new variable's perspective.
• For each limit, substitute the original variable with the terms defined in your substitution equation.

In our example, after setting \( u = x-3 \):

• When \( x = 4 \), replace it in the equation so \( u = 4 - 3 = 1 \).
• When \( x = 9 \), do the same to get \( u = 9 - 3 = 6 \).

This changes the integral limits to \( u = 1 \) and \( u = 6 \).
Adjusting limits ensures that you're evaluating the integral over the correct range in the simplified form. Always remember to transform limits whenever you substitute variables.

Integration by Substitution

Integration by substitution is a powerful technique often employed when the function to be integrated is a composite or a complex rational expression. Here's how it typically proceeds:

• First, execute a substitution that simplifies the components of the integrand.
• Rewrite the entire integrand using the substitution, which might transform parts like fractions into simpler forms.

Continuing our example, transforming \( \int_{4}^{9} \frac{x}{x-3} \, dx \) with \( u = x-3 \) modifies the expression to \( \int_{1}^{6} \left( 1 + \frac{3}{u} \right) \, du \).

• Separate the integrand into manageable parts, like \( \int_{1}^{6} 1 \, du \) and \( 3 \int_{1}^{6} \frac{1}{u} \, du \).
• Individually solve each integral, applying appropriate rules.

This technique is efficient in breaking down complex integrals into simpler, solvable steps and is broadly useful due to its general applicability.