Problem 7 - Calculus

Question

Find the derivatives of the following functions \(F(x)\). \(\int_{x}^{x+1} v(t) d t\) (a "running average" of \(v\) )

Step-by-Step Solution

Answer

The derivative is \( F'(x) = v(x+1) - v(x) \).

Step 1: Understand the Problem

We need to find the derivative of the function \( F(x) = \int_{x}^{x+1} v(t) \, dt \). This is an integral with variable limits of integration, which represents the accumulated value of \( v(t) \) from \( x \) to \( x+1 \).

Step 2: Apply the Leibniz Rule

The Leibniz rule allows us to differentiate an integral with variable limits. The rule states that if \( F(x) = \int_{g(x)}^{h(x)} f(t) \, dt \), then the derivative \( F'(x) \) is given by \( F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x) \).

Step 3: Identify Functions and Limits

In our integral \( F(x) = \int_{x}^{x+1} v(t) \, dt \), we have \( g(x) = x \) and \( h(x) = x+1 \), both of which have derivatives of \( g'(x) = 1 \) and \( h'(x) = 1 \), respectively.

Step 4: Differentiate Using Leibniz Rule

Applying the Leibniz rule, we have:\[ F'(x) = v(h(x)) \cdot h'(x) - v(g(x)) \cdot g'(x) \]Substituting the values:\[ F'(x) = v(x+1) \cdot 1 - v(x) \cdot 1 = v(x+1) - v(x) \].

Step 5: Final Step: Write Down the Result

The derivative of the function \( F(x) = \int_{x}^{x+1} v(t) \, dt \) with respect to \( x \) is \( F'(x) = v(x+1) - v(x) \).

Key Concepts

Derivative of Integral

When working with integrals, it's crucial to understand that they represent accumulation, while derivatives give us rates of change. In this problem, we begin with the function:

• \( F(x) = \int_{x}^{x+1} v(t) \, dt \)

This integral tells us the total change of the function \( v(t) \) over the interval from \( x \) to \( x+1 \). To find the derivative \( F'(x) \), we can employ the Leibniz Rule, which is particularly useful when the limits of integration are functions of \( x \).
By differentiating this integral, we compute how \( F(x) \) changes as \( x \) changes. Here, the derivative indicates the difference in the values of \( v(t) \) at the boundaries \( x \) and \( x+1 \) as expressed by the result:

• \[ F'(x) = v(x+1) - v(x) \]

This expression shows that the change in the running average, \( F(x) \), between those points is the actual difference in the function values at the upper and lower limits. A deep understanding of derivatives extends beyond simple functions to grasp how functions change over intervals and how this impacts the overall rate of change.

Variable Limits of Integration

In this problem, we're dealing with an integral that has variable limits. Specifically, the integral

• \( \int_{x}^{x+1} v(t) \, dt \)

The limits of integration, \( x \) and \( x+1 \), depend on the variable \( x \). This is different from a definite integral, which has constant boundaries. Dealing with variable limits introduces a layer of complexity that calls for special methods to differentiate such integrals.
The Leibniz Rule is suitable here because it helps us find the derivative by accounting for the change in both the function \( v(t) \) and its limits. The use of Leibniz Rule is particularly valuable whenever the boundaries of the integration are not fixed values. By implementing this rule, one can determine how the function behaves at these changing limits and derive accurate information about the function's rate of change.

Differentiation Techniques

Differentiating integrals with variable limits requires specific strategies. The Leibniz Rule offers a way to achieve this by differentiating the integral:

• If \( F(x) = \int_{g(x)}^{h(x)} f(t) \, dt \)
• Then, \( F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x) \)

This formula shows that we calculate the derivative by evaluating our integrated function at its limits and multiplying by the derivatives of those limits.
For the problem given:

• \[ h(x) = x+1 \quad \text{and} \quad g(x) = x \]
• Their derivatives are \( h'(x) = 1 \) and \( g'(x) = 1 \)

Incorporating these, the derivative becomes:

• \[ F'(x) = v(x+1) \cdot 1 - v(x) \cdot 1 = v(x+1) - v(x) \]

This solution highlights the importance of applying appropriate differentiation techniques. By understanding and applying the right rule, you can handle even complex integrals effectively.