Problem 16

Question

In Problems 15-34, use the method of substitution to find each of the following indefinite integrals. $\int \sqrt[3]{2 x-4} d x$

Step-by-Step Solution

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Answer

$\frac{3}{8}(2x - 4)^{4/3} + C$

1Step 1: Choose a substitution

To simplify the integral, let's choose a substitution that can help eliminate the cube root. Choose $u = 2x - 4$.

2Step 2: Differentiate the substitution

Differentiate $u = 2x - 4$ with respect to $x$ to find $du$. This gives $du = 2 \, dx$.

3Step 3: Solve for dx

Rearrange the expression for $du$ to solve for $dx$. This gives us $dx = \frac{du}{2}$.

4Step 4: Substitute into the integral

Replace $2x - 4$ with $u$ and $dx$ with $\frac{du}{2}$ in the original integral. The integral becomes $\int \sqrt[3]{u} \cdot \frac{du}{2}$.

5Step 5: Simplify the integral

Factor out the constant \ \frac{1}{2} \ from the integral, giving $\frac{1}{2} \int u^{1/3} \, du$.

6Step 6: Integrate with respect to u

Using the power rule for integration, integrate $u^{1/3}$ to get $\frac{3}{4}u^{4/3} + C$ where $C$ is the constant of integration.

7Step 7: Substitute back for x

Substitute $u = 2x - 4$ back into the integral result to express the answer in terms of $x$. This gives $\frac{3}{8}(2x - 4)^{4/3} + C$.

Key Concepts

Substitution MethodIntegration TechniquesCube Root Integration

Substitution Method

The substitution method is an elegant and useful technique to simplify finding indefinite integrals. The goal is to simplify the integral into a form that is easier to work with, by introducing a new variable, often denoted as $ u $. This is particularly helpful when dealing with composite functions or complex expressions. It essentially works like a change of variables.

The first step is to identify a piece within the integral that can be substituted. In our problem, we identified $ 2x - 4 $ as a suitable candidate.
Next, we differentiate the substitution, $ u = 2x - 4 $, to find $ du $. This allows us to express $ dx $ in terms of $ du $.
Once the substitution and its derivative have been identified, we replace the chosen expression and $ dx $ in the integral with $ u $ and $ \frac{du}{dx} $ respectively.
With this substitution, the integral in terms of $ x $ turns into a simpler integral in terms of $ u $, which we can then integrate normally.

Finally, don't forget to substitute back the original expression for $ u $ in terms of $ x $ to find the initial integral's solution. It's like unwinding the substitution to give the full answer.

Integration Techniques

Numerous integration techniques build our toolkit for finding indefinite integrals effectively. Among the most common are:

Basic integration rules, such as power, constant, and sum rules.
Integration by substitution (as illustrated in our exercise) is a technique that transforms challenging integrals into more manageable forms.
Integration by parts, useful when dealing with the product of two functions, relies on applying the product rule in reverse.
The method of partial fractions, which is especially useful with rational functions, breaks fractions into a sum of simpler terms.

Each technique has its advantages in different scenarios. Substitution is particularly versatile when dealing with integrals of composite functions or where you can identify a derivative within the integral. When deciding which technique to use, consider the form of the integral; seeing patterns or recognizable structures is key. With practice, you'll automatically sense which method fits best, making integration a much smoother process.

Cube Root Integration

Cube root integration can be intimidating at first, but with the right approach, it becomes approachable. Cube roots involve expressions of the form $ \sqrt[3]{x} $, and integrating such expressions often requires tweaking and simplification.In the provided exercise, the cube root $ \sqrt[3]{2x - 4} $ was simplified using substitution. The integration process here follows a straightforward path once the substitution has been made:

After the substitution, the cube root becomes $ u^{1/3} $, a simple power function in terms of $ u $.
This conversion opens the door to applying the power rule of integration, $ \int u^n \, du = \frac{u^{n+1}}{n+1} + C $, making it easier to integrate.
After simplification and integration, always remember to express the result back in terms of the original variable.

This method emphasizes the utility of substitution for dealing with cube roots, transforming complex roots into something more tractable. With a consistent approach, even tricky integrals become simple to resolve, letting us explore more challenging mathematical terrain with confidence.

Problem 16

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Problem 15

Suppose that $\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1$, and $\int_{0}^{2} g(x) d x=4$. Use properties of definite integra

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Problem 16

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Problem 16

Suppose that $\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1$, and $\int_{0}^{2} g(x) d x=4$. Use properties of definite integra

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Problem 17

In Problems 17-22, calculate $\int_{a}^{b} f(x) d x$, where a and $b$ are the left and right end points for which $f$ is defined, by using the Interval Ad

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