Problem 94 - 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). $$ \int \frac{x}{\sqrt[3]{x+1}} d x $$

Step-by-Step Solution

Answer

The integrated expression is \( \frac{3}{5} (x+1)^{5/3} - \frac{3}{2} (x+1)^{2/3} + C \).

Step 1: Choose Substitution

To solve the given integral using substitution, we first choose an appropriate substitution. As suggested, we let \( u = x + 1 \). This implies that \( x = u - 1 \) and \( dx = du \).

Step 2: Rewrite the Integral

Substitute \( u = x + 1 \) and \( x = u - 1 \) into the integral. The integral becomes:\[ \int \frac{x}{\sqrt[3]{x+1}} \, dx = \int \frac{u-1}{\sqrt[3]{u}} \, du \]

Step 3: Simplify the Expression

Simplify the expression inside the integral:\[ \int \frac{(u-1)}{\sqrt[3]{u}} \, du = \int (u^{2/3} - u^{-1/3}) \, du \]

Step 4: Integrate Each Term

Integrate each term separately:1. \( \int u^{2/3} \, du = \frac{3}{5} u^{5/3} \)2. \( \int -u^{-1/3} \, du = -\frac{3}{2} u^{2/3} \)Combine the results:\[ \frac{3}{5} u^{5/3} - \frac{3}{2} u^{2/3} + C \]

Step 5: Substitute Back to Original Variable

Substitute back the original variable \( x \) by replacing \( u \) with \( x+1 \):\[ \frac{3}{5} (x+1)^{5/3} - \frac{3}{2} (x+1)^{2/3} + C \]

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration. Integration can be thought of as the opposite of differentiation. Where differentiation deals with rates of change, integration helps in finding the total or accumulated quantities. It is widely used to calculate areas under curves, volumes, and much more.

There are mainly two types of integrals:

• Indefinite Integrals: These represent a family of functions and include an arbitrary constant of integration, denoted as "C".
• Definite Integrals: These calculate the net area under a curve from one point to another, and give a specific number as a result.

Understanding integral calculus can open doors to solving more complex mathematical problems, especially in physics and engineering. Whether you are calculating the distance traveled by a car or the amount of water flowing from a tap, integrals provide the tools needed to solve such problems efficiently.

Substitution Method

The substitution method is a technique used to make integration simpler. It's especially useful when dealing with complex integrals that don’t easily fit into standard forms. By using substitution, you change variables to convert the integral into a more manageable form.

In essence, substitution involves:

• Choosing a new variable "u" based on a part of the original equation. A common guideline is to select the inner function or the one that appears in a nested format.
• The derivative of this new function \(du\) should also relate to \(dx\). This makes it easy to replace all occurrences of \(x\) and \(dx\) in terms of \(u\) and \(du\).
• After computing the integral in terms of \(u\), substitute back to the original variable to find the final answer.

This method simplifies the integration process, helping you deal with functions involving powers, compositions, and composite functions more effectively.

Definite Integrals

Definite integrals are extensions of indefinite integrals, used to compute the area under curves between two points. Unlike indefinite integrals, they provide a specific numerical result. This gives definite integrals a substantial application in real-world problems, such as calculating total income over a period or the distance traveled.

Basic steps for solving definite integrals include:

• Identify the limits of integration, usually denoted as \(a\) and \(b\), where the integral evaluates the function from \(a\) to \(b\).
• After performing the integration, apply the Fundamental Theorem of Calculus, which involves substituting the limits into the antiderivative to calculate the difference.
• In some integration techniques, like substitution, make sure to adjust limits if you change variables during the process, or revert back to the original variable once integrated.

Definite integrals are arguably more powerful than indefinite integrals, providing concrete solutions to accumulation problems, whether it involves areas, volumes, or other quantities.