Problem 19 - Applied Calculus

Question

Use a calculator or computer to evaluate the integral. $$\int_{-1}^{1} \frac{1}{e^{t}} d t$$

Step-by-Step Solution

Answer

The value of the integral is approximately 2.35040.

Step 1: Identify the Integral

We are given the integral \( \int_{-1}^{1} \frac{1}{e^t} \, dt \). This is a definite integral that will be evaluated from \( t = -1 \) to \( t = 1 \).

Step 2: Simplify the Integrand

The integrand \( \frac{1}{e^t} \) can be rewritten as \( e^{-t} \). Therefore, the integral to evaluate is \( \int_{-1}^{1} e^{-t} \, dt \).

Step 3: Evaluate the Indefinite Integral

The indefinite integral of \( e^{-t} \) is \( -e^{-t} \). This can be confirmed by differentiating \( -e^{-t} \) which results in \( e^{-t} \).

Step 4: Apply the Limits

Using the Fundamental Theorem of Calculus, evaluate \( -e^{-t} \) from \( t = -1 \) to \( t = 1 \): \[ \left. -e^{-t} \right|_{-1}^{1} = -e^{-1} - (-e^{1}) \]

Step 5: Simplify the Expression

Compute the result by simplifying the expression: \[ -\frac{1}{e} + e = e - \frac{1}{e} \].

Step 6: Use a Calculator

Calculate \( e - \frac{1}{e} \) using a calculator. Approximate \( e \approx 2.71828 \), thus: \[ 2.71828 - \frac{1}{2.71828} \approx 2.71828 - 0.36788 \approx 2.35040 \].

Key Concepts

Definite Integral

A definite integral is a way to calculate the area under a curve between two limits, or boundaries. The integral symbol \( \int \) with the limits \( a \) and \( b \) at the bottom and top, respectively, tells you it's a definite integral. It represents the total accumulation of quantities, such as areas under curves. In the exercise, the definite integral we need to evaluate is \( \int_{-1}^{1} \frac{1}{e^t} \, dt \). Here, the curve is represented by the function \( \frac{1}{e^t} \), and the boundaries are from \( t = -1 \) to \( t = 1 \). By evaluating this definite integral, we are finding the total area under the curve of \( e^{-t} \) from \(-1\) to \(1\).

Definite integrals are very common in calculus because they provide concrete values, unlike indefinite integrals, which often result in a function plus a constant. They're widely used to calculate things like distances, probabilities, and other quantities that can be interpreted as areas.

Exponential Functions

Exponential functions, like the one we encountered in this integration problem, take the form \( a^x \), where \( a \) is a positive constant, and \( x \) is the exponent. In our case, the function \( e^t \) can be seen as the base of the natural logarithm \( e \) raised to the power of \( t \). These functions grow rapidly as the exponent increases. Their inverses also decrease very quickly, making them unique in mathematical modeling.

The reciprocal function \( \frac{1}{e^t} \) simplifies to \( e^{-t} \), which is the integrand we used. Such a transformation is vital because it illustrates the property of exponential functions where the reciprocal becomes an exponent in the negative direction. The behavior of exponential functions, specifically with base \( e \), makes them crucial in disciplines ranging from finance to physics for modeling growth processes.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration, two core operations in calculus. This theorem tells us how to evaluate a definite integral when we know an antiderivative of the function. If a function \( f(t) \) is continuous over an interval \([a, b]\), and \( F(t) \) is its antiderivative, the theorem states that:\[\int_{a}^{b} f(t) \, dt = F(b) - F(a).\]

In our problem, the function \( f(t) = e^{-t} \) has an antiderivative \( F(t) = -e^{-t} \). By applying the theorem, we substitute the limits into this antiderivative: \( F(1) = -e^{-1} \) and \( F(-1) = -e^1 \), then subtract to find the result. This method greatly simplifies finding the area under complex curves, and its application was crucial in arriving at the final result of our integral.