Problem 2
Question
Use the first part of the Fundamental Theorem of Calculus together with the ideas of Examples 1 and 2 to evaluate the definite integrals in Exercises \(1-10\) $$ \int_{-1}^{0} e^{-x} d x $$
Step-by-Step Solution
Verified Answer
The value of the definite integral is \( e - 1 \).
1Step 1: Recognize the Integral Form
The given integral is \( \int_{-1}^{0} e^{-x} \, dx \). The integral is in the form \( \int_{a}^{b} f(x) \, dx \), where \( f(x) = e^{-x} \), \( a = -1 \), and \( b = 0 \).
2Step 2: Identify the Antiderivative
To evaluate the integral using the Fundamental Theorem of Calculus, we need the antiderivative of \( f(x) = e^{-x} \). The antiderivative of \( e^{-x} \) is \( -e^{-x} \).
3Step 3: Apply the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(x) = -e^{-x} \), so we calculate \( F(0) - F(-1) \).
4Step 4: Evaluate at the Upper Limit
Calculate \( F(0) = -e^{-0} = -1 \).
5Step 5: Evaluate at the Lower Limit
Calculate \( F(-1) = -e^{-(-1)} = -e^{1} = -e \).
6Step 6: Calculate the Definite Integral
Substitute the values into the theorem: \( F(0) - F(-1) = -1 - (-e) = -1 + e \).
7Step 7: Simplify the Result
The final result is \( e - 1 \).
Key Concepts
Definite IntegralsAntiderivativesExponential Functions
Definite Integrals
Definite integrals help us find the accumulated area under a curve between two specific points on a graph. This is a core aspect of calculus, allowing for the calculation of total accumulated quantities. A definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits respectively. This setup tells us the interval over which we are finding the area.
This area takes into account the shape and slope of the curve created by the function \( f(x) \) between points \( a \) and \( b \).
This area takes into account the shape and slope of the curve created by the function \( f(x) \) between points \( a \) and \( b \).
- The value may be positive, negative, or zero.
- Positive if the area is above the x-axis, and negative if below.
- Zero can indicate a balance of areas above and below the x-axis.
Antiderivatives
In calculus, an antiderivative is a function whose derivative gives the original function back. Essentially, it's the reverse process of differentiation. When working with definite integrals, finding the antiderivative is crucial. For example, if \( f(x) = e^{-x} \), then its antiderivative is \( F(x) = -e^{-x} \). The fundamental theorem of calculus tells us how to use antiderivatives to evaluate integrals.
- Given \( F(x) \) as an antiderivative of \( f(x) \), the definite integral \( \int_{a}^{b} f(x) \, dx \) equals \( F(b) - F(a) \).
- This provides a simple method to compute areas or accumulated changes quickly.
Exponential Functions
Exponential functions are functions of the form \( f(x) = a^{x} \), where the base \( a \) is a constant and the exponent \( x \) is a variable. They grow or decay at rates proportional to their current value, making them applicable in modeling real-world phenomena like population growth and radioactive decay.
In mathematics, the function \( e^{x} \) is a specific exponential function where the base \( e \) is Euler's number, approximately 2.718. This function exhibits unique properties:
In mathematics, the function \( e^{x} \) is a specific exponential function where the base \( e \) is Euler's number, approximately 2.718. This function exhibits unique properties:
- The derivative of \( e^{x} \) is \( e^{x} \); it remains unchanged upon differentiation.
- Conversely, the antiderivative of \( e^{x} \) is also \( e^{x} \).
Other exercises in this chapter
Problem 2
A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and
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Determine a substitution that will simplify the integral. In each problem, record your choice of \(u\) and the resulting expression for \(d u .\) Then evaluate
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Suppose that \(\int_{2}^{12} g(x) d x=-6\) and \(\int_{2}^{6} g(x) d x=-12\). Evaluate \(\int_{6}^{12} g(x) d x\).
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P In Exercises \(1-10\), write out the sum, and perform the addition. $$ \sum_{j=0}^{6}(2 j-1) $$
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