Problem 274 - Calculus Volume 1

Question

In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int(7 x-11)^{4} d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(\frac{(7x - 11)^5}{35} + C\).

Step 1: Identify a Suitable Substitution

The function within the integral \(7x - 11\) is raised to the power of 4. This suggests using a substitution to simplify the integrand. Let \(u = 7x - 11\). Then, the differential \(du = 7 dx\).

Step 2: Solve for dx in Terms of du

From the equation \(du = 7 dx\), solve for \(dx\) to facilitate the substitution: \(dx = \frac{du}{7}\).

Step 3: Substitute and Simplify the Integral

Substitute \(u = 7x - 11\) and \(dx = \frac{du}{7}\) in the original integral. This gives: \(\int (u)^4 \frac{du}{7} = \frac{1}{7} \int u^4 du\)

Step 4: Integrate the Simplified Expression

Integrate \(u^4\) with respect to \(u\): \(\int u^4 \, du = \frac{u^5}{5} + C\), where \(C\) is the constant of integration.

Step 5: Multiply Back the Constant

Multiply the result by \(\frac{1}{7}\) which was factored out initially: \(\frac{1}{7} \cdot \frac{u^5}{5} = \frac{u^5}{35} + C\)

Step 6: Back-Substitute the Original Variable

Replace \(u\) with \(7x - 11\) to express the integral in terms of \(x\): \(\frac{(7x - 11)^5}{35} + C\)

Key Concepts

Integration by Substitution

Integration by substitution is a technique used to simplify the integration process. It's especially useful when dealing with composite functions, like when a term is raised to a power, as seen in our exercise. The core idea is to replace a complicated part of the integrand with a single variable. This method often makes integration by standard formulas possible.

In the provided exercise, the expression \((7x - 11)^4\) shows a form that could benefit from substitution. By letting \(u = 7x - 11\), we simplify the integrand, turning it into a polynomial in terms of \(u\). This substitution transforms the original integral into a more straightforward problem.

Here's a quick recap:

• Choose \(u\) to replace a part of the integrand that complicates direct integration.
• Differentiate \(u\) to find \(du\). This helps in expressing \(dx\) in terms of \(du\).
• Substitute into the integral, aiming for simpler integration.

Successfully applying substitution effectively reduces complex expressions, as seen in this problem, resulting in easier calculations.

Calculus Problem Solving

Solving calculus problems, like indefinite integrals, often involves recognizing patterns and utilizing strategies like substitution for simplification. The key is analytical thinking and systematic approaches. This means:

• Identifying familiar forms: often expressions contain nested functions which suggest methods like substitution.
• Solving step-by-step: break the entire process into manageable parts. Each step logically follows the last, making the entire process coherent.
• Back-substituting: once the integral is solved in terms of the substitution variable, re-express it in the original terms of the problem.

The problem provided is a classic example of these strategies. We identify the component \(7x - 11\), substitute it with \(u\), simplify, integrate, and finally revert to the original variable. This stepwise method is crucial for handling complex calculus problems.

Change of Variables in Integration

Change of variables is a fundamental concept in calculus, pivotal for dealing with integrals that aren't initially obvious to solve. This technique involves replacing variables with new ones to transform the integral into a more workable form.

In the integral \(\int(7x-11)^4 \, dx\), changing the variable to \(u = 7x - 11\) serves two main purposes:

• It simplifies the function: breaking down a **complex integrand** into a simple algebraic form.
• It rewires the perspective of integration: allows the integral to be understood in simpler terms, which is easier to integrate especially when functions involve powers or other operations.

After substitution, the integration becomes about handling a straightforward power of \(u\).

The final step always involves translating back from the substitute (here, \(u\)) to the original variable, providing the solution in a meaningful and usable form in the context of the initial problem setup. Mastering this concept enables tackling a wide range of otherwise difficult integrals with increased confidence and skill.