Problem 14 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$= f(t)=e^{-t} \quad a=-1 $$

Step-by-Step Solution

Answer

\( F(x) = e - e^{-x} \).

Step 1: Understanding the Problem

We need to find the function \( F(x) \) represented by the integral \( \int_{a}^{x} f(t) \, dt \) where \( f(t) = e^{-t} \) and \( a = -1 \). The integral represents the area under the curve from \( a \) to \( x \).

Step 2: Set Up the Integral

We set up the integral using the limits of integration and the given function: \( F(x) = \int_{-1}^{x} e^{-t} \, dt \).

Step 3: Integrate the Function

The function to integrate is \( e^{-t} \). The integral of \( e^{-t} \) is \( -e^{-t} \). Therefore, \( F(x) = \left[ -e^{-t} \right]_{-1}^{x} \).

Step 4: Evaluate the Definite Integral

Substitute the limits into the integrated expression: \( F(x) = -e^{-x} - (-e^{1}) \). Simplify to get: \( F(x) = -e^{-x} + e^{1} \).

Step 5: Simplify the Expression

The expression for \( F(x) \) simplifies to \( F(x) = e - e^{-x} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.

Key Concepts

Exponential Functions

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. These functions are ubiquitous in nature, describing phenomena such as population growth, radioactive decay, and compound interest.
At the heart of exponential functions is their form:

• General form: \( f(t) = a^t \) where \( a \) is the base and \( t \) is the exponent. A very special case is when the base \( a \) is the number \( e \), which is approximately 2.718.
• The function \( f(t)=e^{-t} \) appears in many scientific fields, most notably in calculus for modeling continuous growth or decay.

The negative exponent \( -t \) in \( f(t)=e^{-t} \) implies a decaying behavior as \( t \) increases. Meaning, as time moves forward, the function's value shrinks exponentially. This decay is due to the inverse relationship described by the negative sign. Understanding exponential functions is vital as it sets the basis for more complex mathematical operations such as integration.

Fundamental Theorem of Calculus

A cornerstone in calculus, the Fundamental Theorem of Calculus bridges the concept of differentiation and integration.
It is described in two parts. The first part explains that if you have a continuous function \( f \) on an interval \([a, b]\), then the function \( F \) defined by

• \( F(x) = \int_{a}^{x} f(t) \, dt \)

is an antiderivative of \( f \). This means that \( F'(x) = f(x) \), suggesting that integration and differentiation are inverse processes. The process is also sometimes referred to as the "Accumulation Function" as it accumulates the area under the curve.
In the exercise, using the Fundamental Theorem of Calculus helps us find the function \( F(x) \) as the integral of \( e^{-t} \) from \( -1 \) to \( x \). It ensures that finding \( F(x) \) gives an expression for the quantity whose rate of change is described by \( e^{-t} \). The second part of the theorem is utilized when evaluating the definite integral from a fixed lower limit \( a \) to any upper limit \( x \). This results in a straightforward calculation by substituting these bounds into the antiderivative.

Integration Techniques

Integration is a foundational technique in calculus used to find areas under curves, among other things. Different strategies exist, depending on the type of function being integrated, whether polynomial, trigonometric, or exponential.
Some useful integration techniques include:

• For exponential functions like \( e^{-t} \), recognize that they often require direct integration.
• The integral of \( e^{-t} \) is straightforward: the integral is \( -e^{-t} \), which stems from the differentiation rules of exponentials.
• Substitution can sometimes help simplify complex integrals or make them easier to recognize.

In our exercise, direct integration of the exponential function is performed because it directly results in \( -e^{-t} \). This arises from a known rule in calculus that states the integral of \( e^{at} \) with respect to \( t \) is \( \frac{1}{a}e^{at} \) when \( a eq 0 \). Recognizing which integration technique to apply in which situation can save time and simplify problem-solving.