Problem 43 - Calculus

Question

If \(f(t)\) is an antiderivative of \(v(t),\) find antiderivatives of (a) \(v(t+3)\) (b) \(v(t)+3\) (c) \(3 v(t)\) (d) \(v(3 t)\)

Step-by-Step Solution

Answer

(a) \(f(t+3) + C\); (b) \(f(t) + 3t + C\); (c) \(3f(t) + C\); (d) \(\frac{1}{3}f(3t) + C\)."

Step 1: Identify Given Information

We are given that \(f(t)\) is an antiderivative of \(v(t)\). This means that \(f'(t) = v(t)\). We need to find antiderivatives for different modifications of \(v(t)\).

Step 2: Solve part (a) - Find Antiderivative of \(v(t+3)\)

The antiderivative of \(v(t+3)\) is a function whose derivative is \(v(t+3)\). By substitution, let \(u = t + 3\). Then the antiderivative is \(\int v(u) \, du = f(u) + C = f(t+3) + C\).

Step 3: Solve part (b) - Find Antiderivative of \(v(t)+3\)

The antiderivative of \(v(t) + 3\) is found by integrating each term separately: \(\int (v(t) + 3) \, dt = \int v(t) \, dt + \int 3 \, dt = f(t) + 3t + C\).

Step 4: Solve part (c) - Find Antiderivative of \(3v(t)\)

The antiderivative of \(3v(t)\) involves a constant multiple. We have: \(\int 3v(t) \, dt = 3 \int v(t) \, dt = 3f(t) + C\).

Step 5: Solve part (d) - Find Antiderivative of \(v(3t)\)

To find the antiderivative of \(v(3t)\), use substitution: let \(u = 3t\), then \(du = 3 \, dt\) or \(dt = \frac{du}{3}\). Thus: \(\int v(3t) \, dt = \frac{1}{3} \int v(u) \, du = \frac{1}{3}f(u) + C = \frac{1}{3}f(3t) + C\).

Key Concepts

Function Transformation

In calculus, function transformation involves altering a function to create a new function. This can include operations like shifting, stretching, compressing, or reflecting graphs of functions. For example, given a function \(v(t)\), if we transform it to \(v(t+3)\), we're effectively shifting the function to the left by 3 units. This technique is particularly useful when finding antiderivatives as it allows the application of the same differentiation and integration rules to transformed functions.
In the case of \(v(t+3)\), we recognized it as a horizontal shift resulting in the same shape on the graph but starting 3 units earlier. By treating \(t+3\) as a new variable, \(u\), integration becomes straightforward, ultimately yielding an antiderivative like \(f(t+3) + C\). This concept of substitution is crucial as it simplifies finding antiderivatives for more complex transformations.

Integration by Substitution

Integration by substitution is an essential technique akin to the chain rule in differentiation but applied in reverse. It simplifies integrals by changing variables, making messy integrals easier to handle. In the context of finding the antiderivative of \(v(t+3)\), we substitute \(u = t + 3\). This substitution transforms the integral into a more straightforward one: \(\int v(u) \, du = f(u) + C\).
The process is simple:

• Identify part of the integral to substitute.
• Express \(dt\) in terms of \(du.\)
• Rewrite the integral in terms of \(u\).
• Integrate, and then substitute back in terms of the original variable.

Through substitution, the antiderivative of a transformed function, like \(v(3t)\), is significantly easier to find, as demonstrated by switching to variables that simplify the integration.

Constant Multiple Rule

The constant multiple rule is a straightforward integration technique that states the integral of a constant multiplied by a function is the constant times the integral of the function itself. Essentially, when given a function like \(3v(t)\), the rule simplifies the integration to: \(\int 3v(t) \, dt = 3\int v(t) \, dt = 3f(t) + C\).
This rule saves time and reduces complexity, ensuring that constants do not interfere with the integration process. It applies equally to any constant and is vital for simplifying calculations across various integrals. Remembering this rule helps in quickly determining the antiderivative while keeping the focus on the function itself.

Integration of Sum

The integration of a sum of functions can be approached by integrating each part separately and then adding the results. This concept is very intuitive: instead of dealing with a complex expression, we can split the integral into simpler parts. For example, to find the antiderivative of \(v(t) + 3\), we handle each component separately:

• \(\int v(t) \, dt = f(t)\)
• \(\int 3 \, dt = 3t\)

Together, the antiderivative becomes \(f(t) + 3t + C\).
This rule illustrates the distributive property of integrals over addition and helps simplify problems by allowing each term to be integrated independently. The concept streamlines the process of integrating complex expressions, making it a valuable tool for solving antiderivatives with multiple components.