Problem 91 - 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). $$ \int x \sqrt[3]{x-4} d x $$

Step-by-Step Solution

Answer

\( \frac{3}{7} (x-4)^{7/3} + 3(x-4)^{4/3} + C \).

Step 1: Identify Substitution

To solve the given integral \( \int x \sqrt[3]{x-4} \, dx \), we will use substitution. Since \( u = x - 4 \), the expression raised to a power inside the integral is a good choice for substitution.

Step 2: Express x in terms of u

Given \( u = x - 4 \), rearrange to express \( x \) in terms of \( u \):\[ x = u + 4. \]

Step 3: Substitute into Integral

Substitute \( u = x - 4 \) and \( x = u + 4 \) into the integral, and since \( dx = du \), we have: \[ \int (u+4) u^{1/3} \, du. \]

Step 4: Expand the Integrand

Multiply \( (u+4) \) with \( u^{1/3} \):\[ \int (u^{4/3} + 4u^{1/3}) \, du. \] This simplifies our integral into two separate integrals.

Step 5: Integrate Each Term Separately

Now, integrate each term separately: \[ \int u^{4/3} \, du = \frac{3}{7} u^{7/3}, \] \[ \int 4u^{1/3} \, du = 4 \times \frac{3}{4} u^{4/3} = 3u^{4/3}. \]

Step 6: Combine the Results

Combine the results of the individual integrals: \[ \frac{3}{7} u^{7/3} + 3u^{4/3} + C. \]

Step 7: Substitute u Back in Terms of x

Substitute back \( u = x - 4 \) to express the solution in terms of \( x \):\[ \frac{3}{7} (x-4)^{7/3} + 3(x-4)^{4/3} + C. \]

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the accumulation of quantities and the areas under curves.

• It involves techniques to find integrals, which represent the accumulated sum of small parts or functions.
• Integrals are a major tool for solving problems related to volume, area, and displacement.

In integral calculus, a crucial tool is the operation of integration, which is the reverse process of differentiation.
This operation is used to find a data function when its rate of change is known. For example, by integrating a velocity function, you can find the position function.
Knowing integral calculus helps in understanding the behavior of physical systems and is widely applied in engineering, physics, and economics.

Integration by Substitution

Integration by substitution is a method that simplifies complex integrals by changing variables.

• This technique focuses on substitution, where we introduce a new variable to simplify the integral.
• It is similar to the reverse chain rule used in differentiation.

In our problem, we used substitution by setting \( u = x - 4 \) to simplify the integral \( \int x \sqrt[3]{x-4} \, dx \).
By choosing \( u \) as the expression inside the root, we transform the integral into an easier form to integrate.
This method is often used when the integrand is a product of a function and its derivative, or when a function is composed inside another related function.

Mathematical Problem-Solving

Mathematical problem-solving involves a series of logical steps to find solutions to given math problems.

• It requires identifying patterns, structures, and different methods to tackle the problem.
• Simplifying the problem into more manageable parts is often necessary.

In the provided exercise, problem-solving begins with identifying the part of the integrand that can be substituted.
After substitution, each part of the equation is addressed separately, as seen in how we split the integrand into two simpler functions \( (u^{4/3}) \) and \( (4u^{1/3}) \).
Combining back all the results to express them in terms of the original variable completes the problem-solving process.

Calculus Techniques

Several calculus techniques help in evaluating integrals and understanding the behavior of functions.

• Each technique has specific scenarios where it is most effective, like substitution method for nested functions.
• Choosing the right technique depends on the form and complexity of the integrand.

The substitution method can simplify integrals significantly, making them easier to solve by transforming complex expressions into basic ones.
Additionally, it is often paired with integration by parts or partial fractions to handle even more complex integrals.
Mastering a range of techniques ensures versatility and efficiency in solving a variety of mathematical problems.