Problem 22 - Calculus

Question

About a company whose expense rate \(t(x)=6-x\) is decreasing. What is the area \(f(x)\) under the line \(t(x)=6-x\) above the interval from \(x\) to 6 ? What is the derivative of this \(f(x)\) ?

Step-by-Step Solution

Answer

The area is \(f(x) = \frac{x^2}{2} - 6x + 18\) and its derivative is \(f'(x) = x - 6\).

Step 1: Define the Problem

We are given a function \(t(x) = 6 - x\) and need to find the area under this line from \(x\) to 6. The function we need to find, \(f(x)\), represents this area.

Step 2: Set Up the Integral

To find the area under the curve \(t(x) = 6 - x\) from \(x\) to 6, we need to compute the definite integral \(\int_{x}^{6}(6 - x)\, dx\).

Step 3: Compute the Antiderivative

Find the antiderivative of the function \(6 - x\). The antiderivative is obtained by integrating each term separately: \(\int 6 \, dx = 6x\) and \(\int -x \, dx = -\frac{x^2}{2}\). So, the antiderivative is \(6x - \frac{x^2}{2}\).

Step 4: Evaluate the Definite Integral

Substitute the limits 6 and \(x\) into the antiderivative: \[ f(x) = \left[6x - \frac{x^2}{2}\right]_{x}^{6} = \left(6(6) - \frac{6^2}{2}\right) - \left(6x - \frac{x^2}{2}\right).\]Calculate this to find the area function \(f(x)\).

Step 5: Simplify the Area Function

Simplify \[f(x) = (36 - 18) - (6x - \frac{x^2}{2}) = 18 - 6x + \frac{x^2}{2}.\]

Step 6: Find the Derivative of \(f(x)\)

Differentiate the area function \(f(x) = 18 - 6x + \frac{x^2}{2}\) to find its derivative. The derivative is \(f'(x) = -6 + x\).

Step 7: Verify the Derivative

Verify that the derivative \(f'(x) = -6 + x\) matches \(-t(x)\), since \(f'(x)\) should be \(\frac{d}{dx}\int_x^6 t(u)du = -t(x)\). This gives \(-6 + x = -(6-x)\), confirming the derivative is correct.

Key Concepts

Definite Integral

In calculus, the concept of a definite integral is fundamental for finding the area under a curve. Specifically, it involves computing the total "net" area between a given function and the x-axis over a closed interval.
To solve for the area under the function \( t(x) = 6 - x \) from some point \( x \) to 6, we use the definite integral \( \int_{x}^{6} (6 - x) \, dx \). This calculation helps us understand the accumulation of values from the starting point \( x \) to the endpoint 6.

• Understanding the limits: The values at the bottom \( x \) and top \( 6 \) of the integral represent the starting and ending points of the area computation.
• Overall, definite integrals help turn continuous growth, decrease, or fluctuation into a tangible value, often representing physical quantities like distance, area, or volume.

Antiderivative

The antiderivative, or the indefinite integral, is the reverse process of differentiation. Through integration, we find a function whose derivative returns the original function we started with.
In the exercise, the task was to compute the antiderivative of \( t(x) = 6 - x \). This involves finding a function \( F(x) \) such that its derivative \( F'(x) \) would equal \( t(x) \).

• For \( 6 \), the antiderivative is \( 6x \).
• For \( -x \), the antiderivative is \( -\frac{x^2}{2} \).

Thus, the complete antiderivative of \( t(x) = 6 - x \) is \( 6x - \frac{x^2}{2} \). This result allows us to integrate any instance of \( t(x) \) over its domain.

Function Area

Calculating the area under a function corresponds to integrating the function over a given interval. This area is the function \( f(x) \) discussed in the problem.
To find \( f(x) \), you evaluate the antiderivative \( 6x - \frac{x^2}{2} \) at the interval's boundaries. In this case, the boundaries are \( x \) and 6.
The expression \( \left[ 6x - \frac{x^2}{2} \right]_{x}^{6} \) gives us the net area as: \( f(x) = (36 - 18) - (6x - \frac{x^2}{2}) = 18 - 6x + \frac{x^2}{2} \).

• This expression captures the cumulative area beneath \( t(x) = 6 - x \) from \( x \) to 6.
• It is an application of the fundamental theorem of calculus, linking differentiation with integration.

Derivative

The derivative concept measures how a function changes as its input changes. In simpler terms, it is the slope or rate of change of a function at any point. Differentiating the area function \( f(x) = 18 - 6x + \frac{x^2}{2} \) helps find how quickly the area under the function is changing.
The differentiation is performed term by term:

• The derivative of \( 18 \), a constant, is \( 0 \).
• The derivative of \( -6x \) is \( -6 \).
• The derivative of \( \frac{x^2}{2} \) is \( x \).

Thus, the derivative \( f'(x) = -6 + x \) describes how the area function \( f(x) \) changes as \( x \) changes. It turns out that \( f'(x) \) is the negative of \( t(x) \), showing the profound relationship between integration and differentiation.