Problem 16 - Calculus Early Transcendentals

Question

Consider the following functions \(f\) and real numbers a (see figure). a. Find and graph the area function \(A(x)=\int_{a}^{x} f(t) d t\) for \(f\) b. Verify that \(A^{\prime}(x)=f(x)\) $$f(t)=2, a=-3$$

Step-by-Step Solution

Answer

Question: Find the area function A(x) of the constant function f(t)=2 from a=-3 to x, graph A(x), and verify that the derivative of the area function, A'(x), is equal to f(x). Answer: The area function of the constant function f(t)=2 from a=-3 to x is A(x) = 2(x + 3). After graphing A(x), we found that its derivative, A'(x), is equal to 2, which verifies that A'(x) = f(x).

Step 1: Find the Area Function \(A(x)\)

To find \(A(x)\), we need to integrate the function \(f(t)\) from \(a\) to \(x\). Since \(f(t) = 2\), we have: $$A(x) = \int_{a}^{x} f(t) dt = \int_{-3}^{x} 2 dt$$

Step 2: Calculate the Indefinite Integral

Since the function is constant, the integral is easy to compute: $$A(x) = \int_{-3}^{x} 2 dt = 2\int_{-3}^{x} dt = 2(t|_{-3}^{x}) = 2(x - (-3))$$

Step 3: Simplify the Area Function

Now, simplify the area function: $$A(x) = 2(x + 3)$$

Step 4: Graph the Area Function

To graph \(A(x) = 2(x + 3)\), plot a line with slope 2 and y-intercept at \((0, 6)\) which passes through points like \((-3, 0)\), \((-2, 2)\), \((-1, 4)\) and so on.

Step 5: Find the Derivative of the Area Function

To verify that \(A^{\prime}(x)=f(x)\), we will differentiate the area function A(x) with respect to x: $$A^{\prime}(x) = \frac{d}{dx} (2(x + 3))$$

Step 6: Calculate the Derivative

Using the power rule for derivatives, we find that: $$A^{\prime}(x) = 2 * 1 = 2$$

Step 7: Verify that the Derivative is Equal to the Function

We computed that \(A^{\prime}(x) = 2\), and our original function \(f(x) = 2\). Since \(A^{\prime}(x)=f(x)\), we have verified the requested condition.

Key Concepts

Area Function

An area function is essentially a way to calculate the area under the curve of a given function over a specific interval. This requires taking an integral. In the context of calculus, the area function denoted by \( A(x) = \int_{a}^{x} f(t) \, dt \), represents the area from a fixed point \( a \) to some variable point \( x \) along the function \( f(t) \).
The fundamental idea is to assess how much "space" the function takes up between these limits. In the exercise provided, where \( f(t) = 2 \), we integrated this constant function over the range \(-3\) to \( x\), resulting in \( A(x) = 2(x + 3) \). This function describes a simple linear graph. The area function effectively translates how changes in \( x \) affect the cumulative area, reflecting it as a function itself.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration. It is composed of two main parts:

• The first part states that if a function \( f \) is continuous over an interval \([a, b] \), and \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
• The second part establishes that the derivative of the integral function \( A(x) = \int_{a}^{x} f(t) \, dt \) is related directly to the function itself, such that \( A'(x) = f(x) \).

In our specific problem, this means that by calculating \( A'(x) = 2 \), we verified that \( A'(x) = f(x) = 2 \). This shows how differentiation can lead back to the original function, completing the cycle of calculus operations.

Definite Integral

A definite integral is a numerical value representing the area under the curve of a function between two designated points on the function's domain. It uses the integral sign accompanied by specific upper and lower bounds, like \( \int_{a}^{b} f(t) \, dt \). This concept is integral to finding the area function, as it quantifies the cumulative "weight" or accumulation across an interval.
The process involves taking a continuous function and slicing it into infinitesimally small partitions over the interval \([a, b] \), adding these pieces up to get the total area.
For a constant function like the one tackled in the exercise, \( f(t) = 2 \) from \( -3 \) to \( x \), the definite integral ends up as \( 2(x + 3) \), a clear indication of how the area grows as \( x \) increases beyond \( -3 \). This makes the definite integral not just an abstract tool but a practical method for real-world computations.