Problem 15 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{0}^{1} e^{-0.2 t} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( 5 - 5e^{-0.2} \).

Step 1 - Identify the Integral

We are given the definite integral \( \int_{0}^{1} e^{-0.2t} \, dt \). Our task is to evaluate this integral using the Fundamental Theorem of Calculus.

Step 2 - Determine the Antiderivative

The first part of the Fundamental Theorem of Calculus tells us that if \( F(t) \) is an antiderivative of \( f(t) \), then \( \int f(t) \, dt = F(t) + C \). For \( e^{-0.2t} \), an antiderivative can be found by observing that the derivative of \( e^{-0.2t} \) is \(-0.2e^{-0.2t} \). Therefore, the antiderivative is \( F(t) = \frac{-1}{0.2} e^{-0.2t} = -5e^{-0.2t} \).

Step 3 - Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(t) \) is an antiderivative of \( f(t) \), then \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \). Here, \( F(t) = -5e^{-0.2t} \), so we need to evaluate \( F(1) - F(0) \).

Step 4 - Evaluate the Antiderivative at Boundaries

Evaluate \( F(t) \) at the boundaries: \[ F(1) = -5e^{-0.2 \times 1} = -5e^{-0.2} \] \[ F(0) = -5e^{-0.2 \times 0} = -5 \times 1 = -5 \]

Step 5 - Compute the Result

Now calculate the definite integral by subtracting the two results: \[ F(1) - F(0) = -5e^{-0.2} - (-5) = -5e^{-0.2} + 5 \]

Step 6 - Simplify the Expression

Rewrite the expression obtained: \[ 5 - 5e^{-0.2} \]. This is the exact value of the definite integral.

Key Concepts

Definite Integral Evaluation

A definite integral provides the net area under a curve from one point to another on a graph. In this case, we are evaluating the integral:

• \( \int_{0}^{1} e^{-0.2t} \, dt \)

The process involves finding the antiderivative of the function and then computing the difference of this antiderivative evaluated at the upper and lower limits. This method is a direct application of the Fundamental Theorem of Calculus, which connects the concept of differentiation with integration.

To evaluate a definite integral using the Fundamental Theorem of Calculus, you need to:

• Find an antiderivative of the function inside the integral.
• Evaluate this antiderivative at the upper limit.
• Evaluate the antiderivative at the lower limit.
• Subtract the value at the lower limit from the value at the upper limit.

This gives you the net area under the curve for the given bounds. This is exactly what we do in this exercise, leading to the expression \( 5 - 5e^{-0.2} \) as the result.

Antiderivative

An antiderivative is a function whose derivative is the original function. For example, consider the function \( e^{-0.2t} \). Its derivative involves adjusting the exponent in response to the chain rule, which is essential for exponential functions involving constants.

Finding the antiderivative is a crucial step in evaluating definite integrals. With \( e^{-0.2t} \), its antiderivative is determined through:

• Recognizing that the derivative of \( e^{kt} \) is \( ke^{kt} \), where \( k \) is a constant. In our case \( k = -0.2 \).
• Observing that the antiderivative involves dividing by the constant from the exponent, resulting in \( F(t) = \frac{-1}{0.2} e^{-0.2t} = -5e^{-0.2t} \).

This function, \( F(t) = -5e^{-0.2t} \), serves as our tool to evaluate definite integrals by applying the Fundamental Theorem of Calculus. This theorem uses the antiderivative to efficiently compute the definite integral's value over a particular interval.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. The standard form is \( a^x \), but usually presented as \( e^x \) when dealing with natural exponential functions, where \( e \) is Euler's number (~2.71828).

In the context of calculus, exponential functions often appear with a negative exponent, such as \( e^{-0.2t} \). They frequently involve changes that grow or decay over time:

• As the exponent decreases negatively (e.g., a time decay factor), the function's value tends towards zero.
• When paired with derivatives or antiderivatives, they lead to shifts by factors of the exponent.

For our integral, \( e^{-0.2t} \), this type demonstrates exponential decay. Its antiderivative accounts for both the base \( e \) and the adjustments needed for the rate of decay. Understanding exponential behavior and how it integrates over specified limits allows us to solve integrals like \( \int_{0}^{1} e^{-0.2t} \, dt \). Moreover, knowing how exponential functions interact with calculus tools is a significant skill in solving differential equations and modeling continuous growth or decay.