Problem 85 - Applied Calculus

Question

\(85-94 .\) 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). $$ \int(x+1)(x-5)^{4} d x $$

Step-by-Step Solution

Answer

\( \frac{1}{6}(x-5)^6 + \frac{6}{5}(x-5)^5 + C \).

Step 1: Choose the substitution

We observe the term \((x-5)^4\). So, let's choose the substitution variable \(u = x - 5\). This allows us to simplify the problem where the derivative is easier to handle.

Step 2: Express x in terms of u

Since \(u = x - 5\), we can express \(x\) as \(x = u + 5\). This transformation allows us to rewrite the integral in terms of \(u\) instead of \(x\).

Step 3: Differentiate to find du

Differentiate \(u = x - 5\) with respect to \(x\), yielding \(\frac{du}{dx} = 1\), which implies \(dx = du\). This substitution will replace \(dx\) in the integral.

Step 4: Substitute into the integral

Substitute \(x = u + 5\) and \(dx = du\) into the integral: \[\int (x+1)(x-5)^4 \, dx = \int ((u+5)+1)u^4 \, du = \int (u+6)u^4 \, du\]. This expansion simplifies the integrand.

Step 5: Expand the integrand

Expand \((u+6)u^4\) to separate terms: \[\int (u^4 + 6u^4) \, du = \int (u^4 + 6u^4) \, du = \int u^5 \, du + \int 6u^4 \, du\]. Now, each term of the expansion can be integrated separately.

Step 6: Integrate

Calculate the integral of each term using power rule:\[\frac{1}{6}u^6 + 6\left(\frac{1}{5}u^5\right) = \frac{1}{6}u^6 + \frac{6}{5}u^5 + C\]. The constants come from applying the power rule: \(\int u^n \, du = \frac{1}{n+1}u^{n+1}\).

Step 7: Back-substitute x for u

Replace \(u\) with \(x - 5\) to return to the variable \(x\): \[\frac{1}{6}(x-5)^6 + \frac{6}{5}(x-5)^5 + C\]. This restores the original variable in the integral.

Key Concepts

integration techniques

Integration techniques are varied methods we use to find the integral of a function, especially when it cannot be integrated directly with simple rules. One such powerful method is the **substitution method**. This technique is particularly handy when dealing with complex integrals, which otherwise do not fit regular integration formulas.

• **Substitution Method**: This involves selecting a new variable, often "u", to transform the integral into a simpler form. This simplifies the integrand, making it easier to calculate.
• **Choosing the Right Substitution**: The key is to pick a substitution that simplifies the expression inside the integral, usually targeting a term raised to a power.
• **Rewriting the Integral**: Once the substitution is made, both the variables and their differentials are rewritten in terms of the new variable "u".

Using this method, we tackle problems by simplifying parts of the integrand through substitution. This not only makes integration possible but often easier to manage within the context of calculus.

power rule in calculus

The power rule in calculus is one of the foundational techniques for differentiating and integrating polynomial functions. This rule makes it straightforward to find derivatives and integrals for any term with a real number exponent.In the context of integration, the power rule helps in calculating antiderivatives by addressing individual terms. The rule states:

• For differentiation: \( \text{If } f(x) = x^n, \text{ then } f'(x) = nx^{n-1}. \)
• For integration: \( \int x^n \, dx = \frac{1}{n+1}x^{n+1} + C, \text{ for } n eq -1. \)

When applying the substitution method as seen in the original exercise, the power rule is used after transforming the integral into simpler terms. Each individual term of the polynomial is integrated individually using the power rule, simplifying the integration of complex functions into manageable steps. This approach greatly facilitates solving integrals that may appear daunting at first glance.

definite and indefinite integrals

Integrals in calculus are divided into two main types: definite and indefinite integrals. Understanding the difference between these two is key to mastering integration.

• **Definite Integrals**: Calculate the area under a curve between two specific points. They are represented with limits of integration, resulting in a real number. \( \int_a^b f(x) \, dx \)
• **Indefinite Integrals**: Represent a family of functions and include a constant of integration "C". They provide the antiderivative of a function. \( \int f(x) \, dx = F(x) + C \)

In the original exercise, the focus is on indefinite integrals. Here, we find the antiderivative of a function without limits, involving a constant "C". This reflects the most basic aspect of the integration process—finding a function whose derivative is the given integrand. Mastery of both definite and indefinite integrals allows for a deeper understanding of areas under curves and the behavior of functions across their domain.