Problem 36 - Calculus

Question

If \(d f / d x=v(x)\) then \(\int v(2 x-t) d x=\) and \(\int v\left(x^{2}\right) x d x=\)

Step-by-Step Solution

Answer

\(\int v(2x-t) dx = \frac{1}{2}f(2x-t) + C\), \(\int v(x^2) x dx = \frac{1}{2}f(x^2) + C\).

Step 1: Understanding the Problem

We need to find the indefinite integrals \(\int v(2x - t) \, dx\) and \(\int v(x^2)x \, dx\). These problems involve substitutions due to the nature of the function inside \(v\).

Step 2: Solving \(\int v(2x - t) \, dx\)

To solve the integral \(\int v(2x - t) \, dx\), we can use a substitution method. Let \(u = 2x - t\). Then, \(du = 2\, dx\). We rearrange this as \(dx = \frac{du}{2}\). Substituting, we get: \[ \int v(u) \cdot \frac{du}{2} = \frac{1}{2} \int v(u) \, du. \] Since \(\frac{df}{du} = v(u)\), \(\int v(u) \, du = f(u) + C\), where \(f(u)\) is the antiderivative of \(v(u)\). Substituting back, \[ \frac{1}{2} f(2x - t) + C. \]

Step 3: Solving \(\int v(x^2) x \, dx\)

For the integral \(\int v(x^2) x \, dx\), let's use another substitution. Let \(u = x^2\). Then, \(du = 2x \, dx\), or \(x \, dx = \frac{du}{2}\). This transforms the integral to: \[ \int v(u) \cdot \frac{du}{2} = \frac{1}{2} \int v(u) \, du. \] Knowing that \(\frac{df}{du} = v(u)\), the solution is \(f(u)\), leading to: \[ \frac{1}{2} f(x^2) + C. \]

Step 4: Summary

We have computed the integrals by substitution. Therefore, \(\int v(2x - t) \, dx\) evaluates to \(\frac{1}{2} f(2x - t) + C\) and \(\int v(x^2) x \, dx\) evaluates to \(\frac{1}{2} f(x^2) + C\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus integration, especially when dealing with complex functions. It simplifies the integration process by transforming the variables, making the integral easier to evaluate.
Think of it as changing the function's face, giving it a new variable name, which allows you to integrate smoothly.

• First, you identify a part of the integral you can substitute. This is typically an inner function, like the terms inside parentheses or exponents.
• Next, you define a substitution variable, often denoted by "u," and express the corresponding "du" in terms of "dx."
• Once the substitution is complete, your integral will change its form to a simpler one, allowing you to proceed with the integration.
• Finally, after integrating, remember to substitute back to the original variable to complete the process.

The substitution makes the integration tractable and often reveals the antiderivative's form quickly.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function that reverses differentiation. Imagine turning back the clock on differentiation to uncover the original function. When we express an antiderivative, we typically add a constant of integration, "C," because there could be many possibilities due to indefinite integration.

• The goal is to find a function "F" such that the derivative of "F" equals the function you are integrating, also called "v" in our problem.
• If the function "v" is the derivative of "F," then the indefinite integral of "v" equals "F(x) + C."
• This process helps us solve equations involving derivatives, giving us the form "F(x)" from "v(x)."

Finding antiderivatives is essential in calculus and is frequently used in solving differential equations and physics problems.

Calculus Integration

Calculus integration refers to the opposite process of differentiation. It's about accumulating a quantity, such as area under a curve. There are two main types: definite and indefinite integrals.

• Indefinite integrals involve finding the antiderivative of a function without specifying limits. These integrate the function across its entire domain.
• Definite integrals calculate the exact accumulation between two points, requiring limits and producing specific values rather than functions.

In practical terms, integration helps in finding areas, volumes, central points, and many others. Whether you're solving physics problems involving motion or business scenarios predicting growth, integration plays a pivotal role.
In our exercises, we focus on indefinite integration, employing the substitution method to simplify the process.

Function Transformation

Function transformation is like reshaping a mathematical function into another form to make solutions or analysis more achievable. We often use this in substitution, when redefining variables in an integral.

• The transformation involves assigning a new variable to a part of the function, simplifying its structure before tackling the main problem.
• For example, by letting "u = 2x - t" or "u = x^2," we turn a more complicated expression into something more straightforward.
• This simplification doesn't change the essence of the problem but allows you to work through it with less difficulty.

Throughout calculus, such transformations help us solve otherwise complex problems neatly, demonstrating their importance in function manipulation and problem-solving.