Question

Variations on the substitution method Evaluate the following integrals. $$\int \frac{x}{\sqrt[3]{x+4}} d x$$

Short answer

Question: Evaluate the integral \(\int \frac{x}{\sqrt[3]{x+4}} d x\). Answer: \(\frac{3}{5}(x+4)^{\frac{5}{3}} - 6(x+4)^{\frac{2}{3}} + C\)

Step-by-step solution

Choose Substitution

To simplify the integral, we can choose the substitution \(u = x + 4\). This means that \(x = u - 4\).

Differentiate Substitution

Differentiate the substitution equation with respect to \(x\) to find \(d x\) in terms of \(d u\). We have: $$ \frac{d u}{d x}=1 \Rightarrow d u = d x $$

Change Variables in the Integral

Substitute \(x\) and \(dx\) in the integral with the corresponding expressions in terms of \(u\): $$ \begin{aligned} \int \frac{x}{\sqrt[3]{x+4}} d x &= \int \frac{u-4}{\sqrt[3]{u}} d u \end{aligned} $$

Rewrite and Separate the Integral

Rewrite and separate the integral as follows: $$ \int \frac{u-4}{\sqrt[3]{u}} d u = \int \left( \frac{u}{\sqrt[3]{u}} - \frac{4}{\sqrt[3]{u}} \right) d u = \int u^{\frac{2}{3}} d u - 4 \int u^{-\frac{1}{3}} d u $$

Integrate Each Expression

Integrate each expression within the integral: $$ \begin{aligned} \int u^{\frac{2}{3}} d u - 4 \int u^{-\frac{1}{3}} d u &= \frac{3}{5} u^{\frac{5}{3}} - 4 \cdot \frac{3}{2} u^{\frac{2}{3}} + C \end{aligned} $$

Replace Back with x

Replace \(u\) with \(x+4\) to obtain the final answer in terms of \(x\): $$ \frac{3}{5}(x+4)^{\frac{5}{3}} - 4 \cdot \frac{3}{2}(x+4)^{\frac{2}{3}} + C $$ So, the integral of the given function is \(\frac{3}{5}(x+4)^{\frac{5}{3}} - 6(x+4)^{\frac{2}{3}} + C\).

Key concepts

Indefinite Integrals

In calculus, an indefinite integral, or antiderivative, is the reverse process of differentiation. It represents the collection of all the antiderivatives of the original function. For instance, if you have a function such as \( f(x) \) and you wish to find \( F(x) \) such that \( F'(x) = f(x) \), then \( F(x) \) is an indefinite integral of \( f(x) \), and can be written with the integral sign as \( \int f(x) \, dx \). As a fundamental concept in calculus, the importance of indefinite integrals exists in their ability to represent accumulated quantities and find original quantities given their rates of change.

For example, if you're given the rate at which a car’s speed is increasing (acceleration), an indefinite integral can tell you the car’s velocity. It's like piecing together a puzzle backward, from the pieces (differential parts) to the completed image (the original function). The constant of integration \( C \) is added to the result because differentiation of a constant is zero; therefore, many functions can have the same derivative and this \( C \) accounts for that unknown constant.

U-substitution

U-substitution is a technique used in integration that simplifies complex integrals by substituting part of the integral with a single variable, usually \( u \). Like a mathematical sleight of hand, this strategy can transform a difficult problem into a more manageable one. When you encounter a function that isn't straightforward to integrate, such as a messy fraction or a composition of functions, u-substitution can often save the day.

The trick is to identify a portion of the integral \( f(x) \) where \( u = g(x) \) can stand in, and finding \( du \) becomes easier. You then express \( dx \) in terms of \( du \) and substitute back into the integral. This substitution process results in an integral that is typically easier to evaluate. Once you find the integral in terms of \( u \) and calculate it, you then substitute back in terms of \( x \) to return to the original variable.

Calculus

Calculus is an elegant and powerful branch of mathematics concerned with the study of change (Differential Calculus) and accumulation (Integral Calculus). It's the mathematical language of motion and growth, providing a framework for understanding the patterns of the universe, from the trajectories of planets to the growth of populations.

The discipline is split mainly into two parts: Differential Calculus, which deals with rates of change, slopes of curves, and the motion of objects, and Integral Calculus, which is all about accumulation, areas under curves, and volumes of solids. The two are tied together by the Fundamental Theorem of Calculus, which shows that differentiation and integration are two sides of the same coin; inverses of each other. Learning calculus opens up a world of problems that can be solved, such as optimizing functions to find maximum profit or minimum cost, understanding rates of reaction in chemistry, or even computing the orbits of objects in space.

Integration Techniques

Integration can be like a puzzle, and just as there’s more than one way to solve a puzzle, there are various techniques to tackle integrals. The main methods include U-substitution, which we discussed above, and it's often the first strategy to try when dealing with a challenging integral. Other techniques include Integration by Parts, which is akin to the product rule for differentiation, and is helpful when you're dealing with the product of two functions. There’s also Partial Fraction Decomposition, a method useful when integrating rational functions.

Beyond these, more advanced methods like Trigonometric Substitution are employed where triangle relationships simplify the integral, and numerical integration methods such as the Trapezoidal Rule and Simpson’s Rule are used when an integral is too complex to solve analytically. Each technique offers its own benefits, and choosing the right one depends on the form of the integral you're working with. Understanding when and how to apply these techniques is crucial for any calculus student and is key to mastering integration.