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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 \(1-10\) $$ \int_{-1}^{0} e^{-x} d x $$

Short Answer

Expert verified
The value of the definite integral is \( e - 1 \).

Step by step solution

01

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 \( b = 0 \).
02

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 \( -e^{-x} \).
03

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 \( F(0) - F(-1) \).
04

Evaluate at the Upper Limit

Calculate \( F(0) = -e^{-0} = -1 \).
05

Evaluate at the Lower Limit

Calculate \( F(-1) = -e^{-(-1)} = -e^{1} = -e \).
06

Calculate the Definite Integral

Substitute the values into the theorem: \( F(0) - F(-1) = -1 - (-e) = -1 + e \).
07

Simplify the Result

The final result is \( e - 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( b \).
  • 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.
The problem given is a definite integral \( \int_{-1}^{0} e^{-x} \, dx \), asking for the total area between \( x = -1 \) and \( x = 0 \), moving along the curve of \( e^{-x} \). These calculations yield an exact area measure for those x-values, illustrating how definite integrals encapsulate the full picture of changes within a specified range.
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.
Understanding and identifying antiderivatives is a key task when solving integrals, making them indispensable for calculating definite integrals.
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:
  • The derivative of \( e^{x} \) is \( e^{x} \); it remains unchanged upon differentiation.
  • Conversely, the antiderivative of \( e^{x} \) is also \( e^{x} \).
However, when the function is \( e^{-x} \), its shape is reflected across the y-axis, demonstrating exponential decay. Calculating the integral of \( e^{-x} \) involves finding the antiderivative \( -e^{-x} \) as in our given problem, crucial in evaluating the definite integral using the fundamental theorem of calculus.

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Most popular questions from this chapter

Simpson's Rule is generally more accurate than the Midpoint Rule, but it is not always more accurate. Calculate \(A=\int_{-1}^{1} \sqrt{|x|} d x .\) With \(N=2,\) estimate \(A\) using both the Midpoint Rule, \(\mathcal{M}_{2},\) and Simpson's Rule, \(\mathcal{S}_{2} .\) What are the absolute errors? Repeat with \(N=4\).

An integral \(\int_{a}^{b} f(x) d x\) and a positive integer \(N\) are given. Compute the exact value of the integral, the Simpson's Rule approximation of order \(N,\) and the absolute error \(\varepsilon\). Then find a value \(c\) in the interval \((a, b)\) such that \(\varepsilon=(b-a)^{5}\left|f^{(4)}(c)\right| /\left(180 \cdot N^{4}\right) .\) (This form of the error, which resembles the Mean Value Theorem, implies inequality \((5.8 .4) .)\) $$ \int_{1}^{4} \sqrt{x} d x \quad N=6 $$

Use Simpson's Rule to estimate cardiac output based on the tabulated readings (with \(t\) in seconds and \(c(t)\) in \(\mathrm{mg} / \mathrm{L}\) ) taken after the injection of \(5 \mathrm{mg}\) of dye. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{t} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \boldsymbol{c}(\boldsymbol{t}) & 0 & 1.9 & 5.8 & 9.4 & 10.4 & 9.1 & 5.9 & 2.1 & 0 \\ \hline \end{array} $$

Calculate \(A=\int_{-1}^{1}|x| d x\). Let \(N\) be an even positive integer that is not divisible by \(4 .\) Show that if a uniform partition of order \(N\) is used, then the Midpoint Rule approximation of \(A\) is exact, but the Simpson's Rule approximation is not.

A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=\ln (x) \quad I=[1,3], N=2 $$

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