Problem 86 - Applied Calculus

Question

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x-2)(x+4)^{5} d x $$

Step-by-Step Solution

Answer

\( \frac{1}{7}(x+4)^7 - (x+4)^6 + C \).

Step 1: Choose the Substitution

Choose the substitution variable to simplify the integral. In this case, let \( u = x + 4 \).

Step 2: Express x in Terms of u

Based on the substitution, express \( x \) in terms of \( u \). Since \( u = x + 4 \), we have \( x = u - 4 \).

Step 3: Differentiate to Find du

Differentiate \( u = x + 4 \) to find \( du \). We get \( du = dx \).

Step 4: Substitute Variables in the Integral

Substitute \( x = u - 4 \) and \( dx = du \) into \( \int(x-2)(x+4)^{5} dx \). The integral becomes \( \int(u-4-2)u^5 du = \int(u-6)u^5 du \).

Step 5: Simplify the Integral Expression

Distribute and simplify the integral expression: \( \int (u^6 - 6u^5) du \).

Step 6: Integrate the Polynomial

Integrate each term separately: \( \int u^6 du = \frac{1}{7}u^7 \) and \( \int 6u^5 du = 6 \cdot \frac{1}{6} u^6 = u^6 \). The integral becomes \( \frac{1}{7}u^7 - u^6 + C \).

Step 7: Substitute Back x for u

Replace \( u \) with \( x + 4 \) in the integrated expression. The solution becomes \( \frac{1}{7}(x+4)^7 - (x+4)^6 + C \).

Key Concepts

Integration Techniques

Integration techniques are methods developed to solve integrals, especially those that aren't immediately solvable with basic formulas. A common method used is the "substitution method." This technique is particularly handy for simplifying integrals by choosing substitutions that make integration more straightforward.

When you use the substitution method, the idea is to identify a part of the integrand (the function being integrated) that can be replaced with a single variable, usually denoted as \( u \). This is especially useful when dealing with composite functions. For example:
For the integral \( \int(x-2)(x+4)^{5} \, dx \), we chose \( u = x + 4 \). This choice simplifies the expression significantly.

Here are some general steps involved in substitution:

• Choose \( u \) for an expression that's raised to a power or makes the derivative straightforward.
• Express other variables in terms of \( u \) using \( x = u - k \).
• Translate \( dx \) into \( du \) by differentiating.

Once the expression is simplified, integrate and then convert back to the original variable.

Calculus Problems

Calculus problems often involve solving complex integrals. These problems can appear daunting at first, but with the right techniques, they become manageable. Many problems in calculus require you to manipulate algebraic expressions and apply specific rules and techniques, such as substitution, to find solutions.

Let's dissect the method applied in our original exercise:

• First, identify parts of the integrand that form a pattern or are repeatedly seen, as these are good candidates for substitution.
• Next, execute algebraic manipulations: Distribute and simplify expressions wherever possible, just as we distributed \( (u-6)u^5 \) to get \( u^6 - 6u^5 \).
• Finally, handle the integration step-by-step by integrating simpler polynomial expressions.

The real challenge is in finding the right approach to tackle each unique problem, often requiring practice to master.

Indefinite Integrals

Indefinite integrals are integrals that do not have specified limits of integration, leading to solutions that include a constant of integration, commonly denoted as \( C \). This constant accounts for all possible vertical shifts of the antiderivative.

In dealing with indefinite integrals, transition from manipulating the function to finding its antiderivative is key.

For instance, consider integrating \( \int(u^6 - 6u^5) \, du \). Each term is integrated separately:

• \( \int u^6 \, du = \frac{1}{7}u^7 \)
• \( \int 6u^5 \, du = 6 \cdot \frac{1}{6}u^6 = u^6 \)

The result is then combined: \( \frac{1}{7}u^7 - u^6 + C \).

Remember to revert the substituted variables back to the original terms to find the complete indefinite integral expression, thus completing the solution.