Problem 54 - Calculus of a Single Variable

Question

Integration as an Accumulation Process In Exercises \(51-54\) , 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

The accumulation function \(F(y) = 8e^{y/2} - 8e^{-1/2}\). The values of \(F(y)\) at \(y=-1, 0, 4\) respectively are \(0\), \(8 - 8/e\), and \(8e^2 - 8/e\).

Step 1: Find the antiderivative

The first step is to find the antiderivative (or the indefinite integral) of the function \(4e^{x/2}\). Here, the rule of exponential function integration \(\int e^{ax} dx = \frac{1}{a} e^{ax}\) is used. Therefore, it's \(\int 4 e^{x/2} dx = 8e^{x/2}\).

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \(F\) is the integral of \(f\) from \(-1\) to \(x\), then \(F'(x) = f(x)\). In this case, \(F(y) = 8e^{y/2}-8e^{-1/2}\) (using the limits \(-1\) and \(y\)).

Step 3: Evaluate \(F(y)\) at \(y=-1\)

Substitute \(y = -1\) into the equation of \(F(y)\), we get \(F(-1) = 8e^{-1/2} - 8e^{-1/2} = 0\)

Step 4: Evaluate \(F(y)\) at \(y=0\)

Substitute \(y = 0\) into the equation of \(F(y)\), we get \(F(0) = 8e^{0} - 8e^{-1/2} = 8 - 8/e\)

Step 5: Evaluate \(F(y)\) at \(y=4\)

Substitute \(y = 4\) into the equation of \(F(y)\), we get \(F(4) = 8e^{4/2} - 8e^{-1/2} = 8e^2 - 8/e\)

Step 6: Graph and interpret result

By graphing the results, the student can visualize the area under the curve that represents the accumulation function.

Key Concepts

Antiderivative

The concept of an antiderivative is critical to understanding the process of integration. Simply put, an antiderivative of a function is another function whose derivative is the original function. For example, if you have a function, let's call it 'f,' and an antiderivative of 'f' is a function 'F,' such that when you take the derivative of 'F,' you get 'f' again. It's like solving a mystery in reverse: you know the effects but are looking for the cause.

Antiderivatives are not unique. This is because the process of differentiation wipes out any constant that might have been part of the original function. That's why when we talk about that antiderivative, we say 'an' antiderivative rather than 'the' antiderivative. Mathematicians denote this by adding a '+ C' to the end of an antiderivative, where 'C' represents any constant. So, in our exercise where the antiderivative of the function is found as an essential part of solving the integral, the constant is not included because the definite integral will evaluate this out.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is the bridge that connects the two main concepts in calculus: differentiation and integration. For students, this theorem is often the 'aha!' moment when the pieces start to fit together. The FTC is built on two parts, but for our exercise, the second part is most relevant. It states that if you have a continuous function 'f' on an interval [a, b] and a function 'F' which is an antiderivative of 'f' on this interval, then the definite integral of 'f' from a to b is equal to 'F(b) - F(a)'.

This equation is incredibly powerful because it gives a way to evaluate definite integrals without the need to find sums of infinite series or areas directly. Instead, if we can find an antiderivative of 'f', we can simply plug in the boundaries of our integral to find the accumulated value. The exercise provided makes practical use of the FTC to evaluate the accumulation function, turning a potentially challenging integral evaluation into a straightforward computation.

Exponential Function Integration

Integrating exponential functions is a common task in calculus, often involving a formula that's almost deceptively simple. Exponential functions, such as the natural exponential function, have the form 'e' raised to a power, and their integrals follow a pattern that correlates closely with their derivatives. In general, the integral of an exponential function can be expressed as \( \int e^{ax} dx = \frac{1}{a} e^{ax} \) plus a constant of integration. However, when dealing with definite integrals, that constant is not needed.

In our exercise, we applied this concept to integrate the function \( 4 e^{x/2} \). By identifying that 'a' in our formula is \(1/2\), we can find the antiderivative to be \( 8e^{x/2} \). Understanding this step is essential, as the ability to integrate exponential functions accurately and confidently will serve students well in various applications, from solving differential equations to analyzing growth and decay problems.

Definite Integral Evaluation

Definite integral evaluation is the process of finding the exact value of an integral between two specific limits. It essentially gives us the net area under the curve of a function on a specified interval. This kind of evaluation moves us from the abstract (finding an antiderivative) to the concrete (calculating a number that has real-world meaning).

In our exercise, we are asked to find the accumulated value represented by the integral of \( 4 e^{x/2} \) from -1 to 'y'. The evaluation process involves substituting the upper limit 'y' and the lower limit -1 into the antiderivative function. Following this, the value from the lower limit is subtracted from the value with the upper limit plugged in, effectively calculating the 'net' area under the curve. By evaluating the integral at different values of 'y', students can understand how changes in these boundaries affect the integral's value, offering a deeper insight into the concept of accumulation in various contexts.