Problem 52 - Essential Calculus

Question

Find the accumulation function \(F\). Then evaluate \(F\) at each value of the independent variable and graphically show the area given by each value of \(F\). $$ F(y)=\int_{-1}^{y} 4 e^{x / 2} d x \quad \text { (a) } F(-1) \quad \text { (b) } F(0) \quad \text { (c) } F(4) $$

Step-by-Step Solution

Answer

Accumulation function: \(8e^{x/2}\). Values at which the function is evaluated: \(F(-1) = 0\), \(F(0) = 8(1 - √(1/e))\), \(F(4) = 8(e^2 - √(1/e))\)

Step 1: Find the Accumulation Function

Let's first find the indefinite integral of the given function, \(4 e^{x / 2}\). This can be done easily using the exponential rule of integration, which states that the integral of \(e^ax\) is \(\frac{1}{a}e^ax\). After integrating we get: \(8e^{x/2}\).

Step 2: Evaluate the Function at F(-1)

Now, substitute [{-1, y}] into the integrated function to get \(F(-1)\). After substitution, we have: \(8e^{-1/2} - 8e^{-1/2} = 0\)

Step 3: Evaluate the Function at F(0)

Now evaluate \(F(0)\). This can be done by substituting [-1, 0]. The solution to this is \(8e^{0} - 8e^{-1/2}\)

Step 4: Evaluate the Function at F(4)

Finally, evaluate \(F(4)\). Substitute these values into the integrated function to get: \(8e^{4/2} - 8e^{-1/2}\)

Step 5: Solving for values

On solving the equations we get the values for \(F(-1)\), \(F(0)\) and \(F(4)\). They are respectively: 0, 8(1 - √(1/e)), and 8(e^2 - √(1/e))

Key Concepts

Indefinite Integral

The concept of an indefinite integral, also known as the antiderivative, is a fundamental building block in calculus. It represents finding the original function from its derivative. In other words, if you know the rate at which something changes, the indefinite integral helps you find the total accumulation over time.
In the given exercise, we are tasked with finding the indefinite integral of the function \(4e^{x/2}\). The process involves applying the exponential rule of integration, which simplifies the problem significantly. According to this rule, the indefinite integral of \(e^{ax}\) is \(\frac{1}{a}e^{ax}\).
This rule allows us to compute the integral of \(4e^{x/2}\) and arrive at the expression: \(8e^{x/2}\). This function \(F(y)\) is our accumulation function and sets the base for evaluating definite integrals later on.

Evaluation of Functions

Evaluating a function means substituting specific values for the variable in a function to determine its output. This is crucial in understanding how a function behaves at different points.
In the exercise, evaluations are performed at three points: \(F(-1)\), \(F(0)\), and \(F(4)\). For each case, we substitute the specific values of \(-1\), \(0\), and \(4\) into the previously derived indefinite integral, \(8e^{x/2}\).

• For \(F(-1)\), substituting gives \(8e^{-1/2} - 8e^{-1/2} = 0\).
• For \(F(0)\), it's \(8e^{0} - 8e^{-1/2}\), yielding a non-zero area under the curve.
• Lastly, \(F(4)\) results in \(8e^{4/2} - 8e^{-1/2}\), displaying a larger area due to the positive exponent on \(e\).

Exponential Integration

Exponential integration is a special case of integration where the function includes an exponential term, such as \(e^{ax}\). These terms appear frequently in growth and decay models.
To perform exponential integration, the rule \(\int e^{ax} \, dx = \frac{1}{a}e^{ax} + C\) is used. Here, \(C\) represents the constant of integration, accounting for the family of functions that differ by a constant.
In the given task, exponential integration was applied to \(4e^{x/2}\). By using the rule, the integral becomes \(8e^{x/2}\) after calculating \(\frac{1}{1/2}\), with the constant term omitted if evaluating a definite integral. This method simplifies solving functions with exponential components, offering insights into the behavior of the function over certain intervals.

Graphical Representation

Graphically representing a function helps in visualizing the concept of integration as the area under a curve. It allows the observer to see how changes in the function's parameters affect its graph.
For the given problem, plotting the accumulation functions \(F(-1)\), \(F(0)\), and \(F(4)\) on a graph provides visual insight. Each evaluated function represents a specific area under the curve of \(8e^{x/2}\) from \(-1\) to \(x\).

• \(F(-1)\) shows no area as it evaluates to zero.
• \(F(0)\) presents a small area starting from \(-1\), highlighting the function's behavior from negative to null growth.
• \(F(4)\) reveals a larger area, illustrating the exponential increase and the consequent area accumulation between \(-1\) and \(4\).

Graphical interpretations aid in understanding the dynamics and results of integrations, especially in the context of growth or decay modeled through exponential functions.