Problem 147 - Calculus Volume 1

Question

Set \(\quad F(x)=\int_{1}^{x}(1-t) d t .\) Find \(\quad F^{\prime}(2)\) and the average value of \(F^{\prime}\) over \([1,2]\)

Step-by-Step Solution

Answer

\( F'(2) = -1 \); average value of \( F'(x) \) over \([1,2]\) is \(-0.5\).

Step 1: Understand the Function and its Derivative

Given the function \( F(x) = \int_{1}^{x} (1-t) \, dt \), the derivative \( F'(x) \) can be found using the Fundamental Theorem of Calculus, which states \( F'(x) = 1 - x \). This is because the derivative of an integral with respect to its upper limit is the integrand evaluated at that upper limit.

Step 2: Calculate F'(2)

Substitute \( x = 2 \) into the derivative \( F'(x) = 1 - x \). Thus, \( F'(2) = 1 - 2 = -1 \).

Step 3: Set Up for Average Value Calculation

The average value of a function \( g(x) \) over an interval \([a,b]\) is given by the formula \( \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \). For \( F'(x) \) over \([1,2]\), the formula becomes \( \frac{1}{2-1} \int_{1}^{2} (1-x) \, dx \).

Step 4: Calculate the Integral for Average Value

To calculate \( \int_{1}^{2} (1-x) \, dx \), integrate the function \( 1-x \). The antiderivative of \( 1-x \) is \( x - \frac{x^2}{2} \). Evaluating from 1 to 2 gives:\[ \left[ x - \frac{x^2}{2} \right]_{1}^{2} = \left( 2 - \frac{4}{2} \right) - \left( 1 - \frac{1}{2} \right) = (2 - 2) - (1 - 0.5) = 0 - 0.5 = -0.5 \].

Step 5: Compute the Average Value

Substitute the result from Step 4 into the average value formula: \( \frac{1}{1}(-0.5) = -0.5 \). This is the average value of \( F'(x) \) over \([1,2]\).

Key Concepts

Derivative of an Integral

When dealing with integrals and derivatives, the Fundamental Theorem of Calculus plays a crucial role. This powerful theorem connects these two concepts beautifully. Here's the idea: if you have a function defined by an integral, like our \( F(x) = \int_{1}^{x} (1-t) \, dt \), the derivative of this function with respect to \( x \) can be quickly found. According to the theorem, this derivative is simply the integrand evaluated at \( x \). So for our specific function: - The derivative, \( F'(x) \), is \( 1-x \), obtained by replacing \( t \) with \( x \) in the integrand \( 1-t \). To find \( F'(2) \), simply plug \( x = 2 \) into \( 1-x \): - \( F'(2) = 1 - 2 = -1 \). The beauty of this theorem is in its simplicity, making the process of finding derivatives of such integrals straightforward and efficient.

Average Value of a Function

The average value of a function over a specific interval is a concept that tells us what the 'typical' value of the function is. For a function \( g(x) \) over an interval \([a, b]\), the formula is: \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \] For \( F'(x) = 1-x \) over the interval \([1, 2]\), the average value calculates as follows: - Set \( g(x) = 1-x \). The integral becomes: \[ \int_{1}^{2} (1-x) \, dx \] Solving this integral involves finding the antiderivative: - \( x - \frac{x^2}{2} \) from \( 1 \) to \( 2 \), resulting in a value of \(-0.5\). Now, apply the average value formula: - \( \frac{1}{1}(-0.5) = -0.5 \). Thus, the average value of \( F'(x) \) over \([1, 2]\) is \(-0.5\). This concept of averaging gives insight into the behavior of \( F'(x) \) over the interval rather than just at a point.

Integrals and Derivatives

Understanding integrals and derivatives forms the foundation of calculus. They are inverse operations, meaning one operation can "undo" the other.An integral accumulation conceptically 'adds up' small quantities to find a total value. It's useful for calculating areas, among other applications. Meanwhile, a derivative represents the rate of change of a function. It effectively "slices" up small changes to understand function behavior better.Exploring these through the Fundamental Theorem of Calculus shows how derivatives of integrals yield the original function's rate of change evaluated at a specific point. The derivative \( F'(x) = 1-x \) of our function \( F(x) = \int_{1}^{x} (1-t) \, dt \) exemplifies this principle, showcasing how each tool complements the other in problem-solving.In calculus, this relationship is harnessed to tackle a wide array of problems, blending the total value insights from integrals with precise change information from derivatives. This union is vital for understanding dynamic systems, physical phenomena, and mathematical relationships.