Problem 18 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{0}^{2}\left(e^{t}-e^{-t}\right) d t$$

Step-by-Step Solution

Answer

The definite integral \(\int_{0}^{2}\left(e^{t}-e^{-t}\right) d t\) is equal to \(e^2 - e^{-2}\).

Step 1: Determine the antiderivative of \(f(t)\)

: So the first thing we need to do is to find the antiderivative of function \(f(t) = e^t - e^{-t}\). Computing the antiderivative of this function can be done by finding the antiderivative of each term separately. The antiderivative of \(e^t\) is simply \(e^t\) (since the derivative of \(e^t\) is \(e^t\)), and the antiderivative of \(e^{-t}\) is \(-e^{-t}\) (since the derivative of \(-e^{-t}\) is \(e^{-t}\)). Thus, the antiderivative of \(f(t)\) is \[F(t) = e^t - e^{-t}.\]

Step 2: Evaluate the antiderivative at the bounds of the integral

: Next, we apply the Fundamental Theorem of Calculus, which states that the definite integral of a function over an interval is equal to the difference of its antiderivative evaluated at the endpoints of the interval. Therefore, we have: \[ \int_{0}^{2} \left(e^t - e^{-t}\right) dt = F(2) - F(0).\] Now, we evaluate the antiderivative at these endpoints: \begin{align*} F(2) &= e^2 - e^{-2}, \\ F(0) &= e^0 - e^{-0} = 1- 1 = 0. \end{align*}

Step 3: Compute the definite integral

: Finally, we can compute the definite integral as the difference of the antiderivative evaluated at the endpoints: \[ \int_{0}^{2} \left(e^t - e^{-t}\right) dt = F(2) - F(0) = (e^2 - e^{-2}) - (0) = e^2 - e^{-2}.\] Hence, the definite integral of the function \(f(t) = e^t - e^{-t}\) over the interval \([0, 2]\) is equal to \(e^2 - e^{-2}\).

Key Concepts

Antiderivative

When dealing with definite integrals, the concept of an antiderivative is crucial. An antiderivative is a function whose derivative yields the original function. In simpler terms, if you have a function \( f(t) \), its antiderivative \( F(t) \) is such that when you differentiate \( F(t) \), you get \( f(t) \) back.

To solve the integral \( \int_{0}^{2} (e^t - e^{-t}) \, dt \), we first determine the antiderivative of each part of our function:

• For the term \( e^t \), its antiderivative is \( e^t \) itself. This is because differentiating \( e^t \) results in \( e^t \), making it its own antiderivative.
• For the term \( e^{-t} \), the antiderivative is \( -e^{-t} \). When you differentiate \( -e^{-t} \), the negative sign cancels the original negative of the exponent, yielding \( e^{-t} \).

The complete antiderivative of \( f(t) = e^t - e^{-t} \) is \( F(t) = e^t - e^{-t} \). By understanding and finding antiderivatives, we set the stage for solving definite integrals using the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in the field of integral calculus, linking differentiation and integration in a profound way. It consists of two main parts, but here we focus on the part that helps with definite integrals.

According to the Fundamental Theorem, if \( F(t) \) is an antiderivative of \( f(t) \) on an interval \([a, b]\), then the definite integral of \( f(t) \) from \( a \) to \( b \) is given by:
\[ \int_{a}^{b} f(t) \, dt = F(b) - F(a). \]

In our example, we found that the antiderivative \( F(t) \) for \( f(t) = e^t - e^{-t} \) is \( e^t - e^{-t} \). Applying the theorem:

• Evaluate \( F(t) \) at the upper limit, \( t = 2 \): \( F(2) = e^2 - e^{-2} \).
• Evaluate \( F(t) \) at the lower limit, \( t = 0 \): \( F(0) = e^0 - e^{0} = 0 \).

Thus, the definite integral \( \int_{0}^{2} (e^t - e^{-t}) \, dt \) becomes \( e^2 - e^{-2} - 0 = e^2 - e^{-2} \). The theorem elegantly simplifies calculating the area under a curve by leveraging antiderivatives.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(t) = a^t \) where \( a \) is a constant. A special case is the natural exponential function \( e^t \), where \( e \) is approximately 2.71828, an irrational constant.

These functions are unique because their rate of growth is proportional to their value. This property leads to some intriguing behaviors, especially in calculus:

• The derivative of \( e^t \) is \( e^t \), making it an ideal case for both differentiation and integration.
• The derivative of \( e^{-t} \) is \(-e^{-t} \). Therefore, when integrating, you must consider the negative sign to find its antiderivative correctly.

Understanding exponential functions is crucial for solving many integrals involving the \( e^x \) type, especially when dealing with growth and decay models. In our exercise, the function \( f(t) = e^t - e^{-t} \) combines these properties, demonstrating how exponential functions behave under the operations of calculus.