Problem 20 - Calculus Early Transcendentals

Question

Consider the following functions \(f\) and real numbers a (see figure). a. Find and graph the area function \(A(x)=\int_{a}^{x} f(t) d t\) b. Verify that \(A^{\prime}(x)=f(x)\) $$f(t)=2 t+5, a=0$$

Step-by-Step Solution

Answer

Question: Find the area function A(x) for the function \(f(t) = 2t + 5\) with \(a = 0\), graph it, and verify that \(A'(x) = f(x)\). Answer: The area function A(x) for the given function is \(A(x) = x^2 + 5x\), and its graph is a parabola opening upwards. It is verified that \(A'(x) = f(x)\), as both expressions are equal to \(2x + 5\).

Step 1: Calculate the integral of f(t)

We start by integrating f(t) with respect to t: $$A(x) = \int_{a}^{x} f(t) dt = \int_{0}^{x} (2t + 5) dt$$

Step 2: Compute the area function A(x)

Now we will compute A(x) by calculating the integral we've found. Use the power rule (\(\int t^{n} dt = \frac{t^{n+1}}{n+1}\)) and the constant rule (\(\int k dt = kt\)): $$A(x) = \int_{0}^{x} (2t + 5) dt = \left[ t^2 + 5t \right]_{0}^{x} = x^2 + 5x - (0^2 + 5 \cdot 0) = x^2+5x$$

Step 3: Graph A(x)

To graph the area function A(x), we can plot the points for different values of x and then draw the curve through them. Some of the points for the quadratic function are: (0, 0), (1, 6), (2, 16), (3, 30), (4, 48), (5, 70) Please graph these points and the curve will appear. The graph is a parabola opening upwards.

Step 4: Compute the derivative of A(x)

Now we will find the derivative of A(x) with respect to x: $$A^{\prime}(x) = \frac{d}{dx}(x^2 + 5x) = 2x + 5$$

Step 5: Verify \(A^{\prime}(x) = f(x)\)

We will now verify that the derivative of A(x) is equal to f(x) by comparing the expressions: $$A^{\prime}(x) = 2x + 5 = f(x)$$ Since we found that the derivative of A(x) and f(x) are the same, \(A^{\prime}(x) = f(x)\) holds true.

Key Concepts

Area function

An area function represents the accumulated area under a curve from a starting point \(a\) to another point \(x\). In calculus, this is expressed as an integral:

• The area function \(A(x)\) is defined as \(A(x) = \int_{a}^{x} f(t) \, dt\).
• The function \(f(t)\) is integrated over the interval from \(a\) to \(x\) to find \(A(x)\).

For the given function \(f(t) = 2t + 5\) and \(a = 0\), the area function becomes \(A(x) = \int_{0}^{x} (2t + 5) \, dt\). This integral is solved using basic integration rules resulting in \(A(x) = x^2 + 5x\).
This quadratic function \(A(x)\), when plotted, is a parabola that opens upwards, demonstrating the accumulation of area under the curve as \(x\) increases.
Calculating the area function is a foundational step in integral calculus, providing insights into how functions accumulate over intervals.

Derivative

The derivative is a fundamental concept in calculus that signifies the rate of change of a function. It is essential for understanding how functions behave. For the function \(A(x) = x^2 + 5x\):

• The derivative \(A'(x)\) of the area function is calculated as \(A'(x) = \frac{d}{dx}(x^2 + 5x)\).
• Applying the power rule, the derivative simplifies to \(2x + 5\).

The derivative \(A'(x)\) represents the slope of the tangent line to the curve \(A(x)\) at any point \(x\). Since \(A'(x) = f(x)\), it confirms that the rate at which the area accumulates (\(A'(x)\)) corresponds to the original function \(f(x)\). This relationship is core to understanding how derivatives reflect instantaneous changes within functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the process of differentiation and integration, offering a profound insight into calculus. It is composed of two main parts:

• The first part states that if \(f\) is continuous on \([a, b]\), and \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_a^b f(x) \, dx = F(b) - F(a)\).
• The second part assures that for any continuous function \(f\), the derivative of its integral is the original function: \(\frac{d}{dx}\left(\int_{a}^{x} f(t) \, dt\right) = f(x)\).

In our context, the theorem confirms that the derivative of the area function \(A(x) = \int_{a}^{x} f(t) \, dt\) returns the function \(f(x)\) itself.
Consequently, observing \(A'(x) = f(x)\) showcases the powerful connection between integration and differentiation. This robust relationship forms the backbone of solving many calculus problems, offering insights into the properties and behaviors of functions through their integrals and derivatives.