Problem 23 - Calculus Early Transcendentals

Question

Use a change of variables to find the following indefinite integrals. Check your work by differentiating. $$\int x^{3}\left(x^{4}+16\right)^{6} d x$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function \(x^3\left(x^4+16\right)^6\) with respect to \(x\). Answer: The indefinite integral is \(\int x^{3}\left(x^{4}+16\right)^{6} dx=\frac{1}{28}(x^4 + 16)^7 + C\).

Step 1: Choose the substitution

Let's choose the substitution of \(u\) for the expression that is raised to a power: \(u=x^4+16\).

Step 2: Find the derivative of the substitution

Now, we need the derivative of \(u\) with respect to \(x\). Differentiate \(u\) with respect to \(x\) to obtain \(du\): $$\frac{du}{dx} = \frac{d}{dx}(x^4+16) = 4x^3$$ Now we can express \(dx\) in terms of \(du\): $$dx = \frac{du}{4x^3}$$

Step 3: Rewrite the integral in terms of \(u\)

Substitute \(u\) and \(dx\) into the original integral: $$\int x^3(u)^{6} \frac{du}{4x^3} = \frac{1}{4} \int u^6 du$$ We've effectively simplified the integral using the change of variables method.

Step 4: Solve the integral

Now, we can easily integrate the function with respect to \(u\): $$\frac{1}{4} \int u^6 du = \frac{1}{4} \cdot \frac{u^7}{7} + C = \frac{1}{28}u^7 + C$$

Step 5: Substitute the original expression for u

Replace \(u\) with the original expression in terms of \(x\): $$\frac{1}{28}(x^4 + 16)^7 + C$$

Step 6: Differentiate to check the solution

To check our work, we need to differentiate the result to see if we get back the original integrand. Differentiate the result with respect to \(x\): $$\frac{d}{dx} \left(\frac{1}{28}(x^4 + 16)^7 \right) = 7 \cdot \frac{1}{28}(x^4+16)^6 \cdot 4x^3 = x^3\left(x^4+16\right)^6$$ Since the result matches the original integrand, our solution is correct. The indefinite integral is: $$\int x^{3}\left(x^{4}+16\right)^{6} dx=\frac{1}{28}(x^4 + 16)^7 + C$$

Key Concepts

U-Substitution

Understanding u-substitution is crucial when dealing with integration. It is a method used to simplify complex integrals by introducing a new variable, typically denoted as 'u'. The goal is to transform the given integral into a simpler form which can be easily integrated.

Let's break down the steps in the context of the exercise: First, identify the inner function that's part of a more complex expression, in this case, it was the term within the brackets \(x^{4}+16\). Next, assign this inner function to 'u', which helps in substituting part of the integral. After finding the derivative of 'u' with respect to 'x', which gives you \(du/dx\), you can then rewrite \(dx\) in terms of \(du\) and \(x\). Finally, substitute back into the integral to get an expression solely in terms of 'u', making it more straightforward to integrate.

Integration Techniques

The calculation of indefinite integrals, which represent families of functions plus an arbitrary constant, utilizes various integration techniques. Some well-known methods are integration by parts, partial fractions, and trigonometric substitution. However, one of the most common is u-substitution, which simplifies the process by reducing complex integrals to basic forms.

When you're faced with an integrand that is a product of a function and its derivative, or when part of the integrand is raised to a power, u-substitution is typically a helpful technique. Remember, the choice of 'u' is not always obvious, and sometimes, more than one substitution is possible. Practice, in this case, certainly makes perfect as it improves one's ability to spot the ideal substitution quicker.

Change of Variables

The change of variables is a broader concept which encompasses u-substitution. By changing variables, we essentially restate the problem in a different 'language', which often times can turn a complex problem into a solvable one.

In the given exercise, the variable change was made from 'x' to 'u', where 'u' represented the troublesome part of the integrand. This not only simplified the equation but in some cases, it makes the integral feasible to solve. When it comes to a change of variables, always remember to account for the relationship between 'dx' and 'du'. A correct transition between the variables is absolutely critical for the solution to be accurate.

Differentiating to Check Integration

After solving an integral using techniques such as u-substitution, it's important to verify the solution. This can be done by differentiating the result and checking if it yields the original integrand.

In the exercise provided, differentiation of the final answer gives us back the integrand we started with. This method acts as a confirmation that the integration process was performed correctly. It also provides a valuable learning loop. By differentiating the result, you not only check your work, but also reinforce understanding how integration and differentiation are inverse processes.