Problem 14 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(13-16,\) verify that \(\int f(u) d u \neq \int f(u) d x\) $$f(u)=u^{2}\( and \)u=x^{5}$$

Step-by-Step Solution

Answer

The integral \( \int f(u) \, du \) evaluates to \( \frac{1}{3}u^3 + C \), whereas the integral \( \int f(u) \, dx \) evaluates to \( \frac{1}{11}x^{11} + C \). Since these results are not equal, it verifies the statement \( \int f(u) \, du \neq \int f(u) \, dx \).

Step 1: Find \( \int f(u) \, du \)

First let's solve \( \int f(u) \, du \). Here, \( f(u) = u^{2} \), so we have \( \int u^2 \, du \). The antiderivative of \( u^2 \) is \( \frac{1}{3}u^3 + C \), where C is the constant of integration. So the solution to \( \int u^2 \, du \) is \( \frac{1}{3}u^3 + C \).

Step 2: Find \( \int f(u) \, dx \)

The function \( f(u) \) is given as \( u^2 \) where \( u = x^5 \). We have to substitute \( x^5 \) in place of \( u \) so the integral becomes \( \int (x^5)^2 \, dx = \int x^{10} \, dx \). The antiderivative of \( x^{10} \) is \( \frac{1}{11}x^{11} + C \), so the solution to \( \int f(u) \, dx \) is \( \frac{1}{11}x^{11} + C \).

Step 3: Compare Results

The solution to \( \int f(u) \, du \) was found as \( \frac{1}{3}u^3 + C \), and the solution to \( \int f(u) \, dx \) was \( \frac{1}{11}x^{11} + C \). These results are not equal, therefore we can verify that \( \int f(u) \, du \neq \int f(u) \, dx \).

Key Concepts

Integration

Integration is one of the cornerstone concepts in calculus. It is the process used to find functions given their derivatives. In simpler terms, if you begin with the slope of a curve, integration helps you uncover the shape and area under that curve.
To perform integration, you usually look for the antiderivative of a function. This means finding a function whose derivative is the original function you started with.
When integrating, we often refer to the "indefinite integral," which isn't limited to specific endpoints on the graph and includes a constant of integration, typically denoted as "+ C." It reflects the family of all possible functions that could have the same derivative.

• Integration finds the antiderivative of a function.
• It's essentially the reverse of differentiation.
• Results in a function set including the constant of integration "+C."

In our exercise, we compared two integrals, showing how the variable of integration affects the outcome. This demonstrated how integrals can differ based on their variable of choice.

Antiderivative

The antiderivative is what you get when you integrate a function. In this context, it's the function that, when differentiated, returns the original function. So, if you have a derivative and you want to "reverse" it back into the function, you find its antiderivative.
Let's take an example from our exercise. We identified the antiderivative of the function \( u^2 \) as \( \frac{1}{3}u^3 + C \). Here, when this antiderivative is differentiated, it brings us back to \( u^2 \).
When obtaining antiderivatives, remember:

• Different functions have different techniques for finding their antiderivatives.
• The constant "+C" signifies an infinite number of solutions, as any constant's derivative is zero.
• Antiderivatives are vital in solving integration problems.

In this exercise, two different antiderivatives were solved, illustrating how changing variables can completely alter solutions.

Substitution Method

The substitution method is a powerful technique in integration used to simplify complex functions. It is particularly useful when dealing with integrals involving composite functions or where the integration seems difficult.
The essence of the substitution method is changing the variable of integration to make the integral easier to solve. You replace a complicated part of the function with a simpler variable, solve the integral, and then revert back to the original variable.
In our exercise, we used substitution by letting \( u = x^5 \) to transform the function \( f(u) = u^2 \) into \( (x^5)^2 \). This substitution simplified the integration from a complex power of a variable to \( x^{10} \).

• Helps solve integrals of composite functions.
• Involves changing of variables for simplification.
• Reverts after integration to match original variables.

This method is crucial for tackling difficult integrals by breaking them down into more manageable parts, demonstrating its practicality in calculus.