Problem 13 - 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 \(A^{\prime}(x)=f(x)\) $$f(t)=5, a=0$$

Step-by-Step Solution

Answer

a. The area function A(x) is found by integrating the given function f(t) from a to x, which in this case is from 0 to x. We find that the area function A(x) is 5x. b. To verify that A'(x) = f(x), we find the derivative of the area function A(x) with respect to x and compare it to f(x). We find that the derivative of A(x), A'(x), is also 5, which is equal to the original function f(x). Therefore, we have verified that A'(x) = f(x).

Step 1: Find the Area Function A(x)

To find the area function, we'll need to integrate \(f(t)=5\) with respect to t, while keeping the bounds from a to x. By doing so, we'll obtain the function \(A(x)\): $$ A(x) = \int_{0}^{x} 5 dt $$ Now, let's integrate: $$ A(x) = 5\int_{0}^{x} dt = 5\left[t\right]_0^x = 5(x-0) = 5x $$ Therefore, our area function A(x) is simply: $$ A(x) = 5x $$

Step 2: Graph the Area Function A(x)

With A(x) derived, now it's time to graph the function. Here, since A(x) is a linear function with a slope of 5, the graph will be a straight line passing through the origin (0,0) and increasing with a slope of 5. Keep in mind that this is simply a visualization of the function and is not explicitly shown in this text format. A graphing calculator or dedicated software should be used to visualize the function.

Step 3: Verify that A'(x) = f(x)

Finally, we'll verify that the derivative of A(x) is equal to f(x). To do this, we'll find the derivative of A(x) with respect to x: $$ A'(x) = \frac{d}{dx}(5x) = 5 $$ Now, we'll compare A'(x) to f(x): $$ A'(x) = 5 $$ and, $$ f(x) = f(t) = 5 $$ As we can see, the derivative of the area function A(x) is equal to the original function, f(x). This verifies that: $$ A'(x) = f(x) $$

Key Concepts

Area Function

An Area Function is a powerful concept in integral calculus as it relates the accumulation of area under a curve. Given a constant function like \(f(t) = 5\), the area function \(A(x)\) essentially calculates the space under the curve from a specified point \(a\) to any point \(x\). To do this, we integrate the function between the specified bounds.
In this exercise, the constant function \(f(t) = 5\) means we aim to find the area of a rectangle with height 5 and base \((x - a)\). The integral \(\int_{a}^{x} 5 \, dt\) yields \(5(x - a)\), which simplifies to \(5x\) when \(a = 0\).

• The formula for the area function, \(A(x) = \int_{0}^{x} 5 \, dt = 5x\), shows that \(A(x)\) is linear, representing a line with a slope of 5.
• This highlights how, for constant functions, the area function is simply a multiplication of the constant by the length of the interval \((x - a)\).

By understanding the area function concept, students can certainly grasp how integral calculus translates areas into functions.

Definite Integral

A Definite Integral is a significant concept in integral calculus, utilized to determine the exact area under a curve between two points. In our exercise, the definite integral is represented by \(\int_{0}^{x} 5 \, dt\). This operation is defined over an interval with specified lower and upper bounds \([a, x]\).
Essentially, the definite integral calculates the net area between the curve and the x-axis, which in this case is a constant, simplified by the concept of an area function.

• With a function such as \(f(t) = 5\), the integration process results in \(5(x - a)\) when evaluated between \(t = 0\) and \(t = x\).
• In our example, the outcome is straightforward due to the constant nature of the function and presents as \(5x\).

Definite integrals offer a foundational tool for recognizing exact measures of area, which becomes intuitive when dealing with linear areas under constant functions.

Derivative Verification

Derivative Verification is a procedure used to confirm that the derivative of an area function recovers the original function. This stems from the Fundamental Theorem of Calculus, which states the integral's derivative yields the function we started with.
In this exercise, our area function is \(A(x) = 5x\), and checking its derivative \(A'(x)\) should take us back to \(f(x) = 5\).

• Taking the derivative \(A'(x) = \frac{d}{dx}(5x)\) results in 5, which matches our original function \(f(x)\).
• This validation demonstrates how integration and differentiation are inverse processes, tightly connected through calculus.

By performing this check, students gain confidence and understand that derivatives confirm the accuracy of their integration tasks, cementing their understanding of the entire calculus operation.