/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Evaluate the given definite inte... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{0}^{1} e^{x} d x $$

Short Answer

Expert verified
The definite integral \( \int_{0}^{1} e^{x} \, dx \) evaluates to \( e - 1 \).

Step by step solution

01

Identify the Integrand

The integrand in the problem is the function \( f(x) = e^x \). We need to find the antiderivative of this function in order to evaluate the definite integral \( \int_0^1 e^x \, dx \).
02

Find the Antiderivative

The function \( e^x \) is its own antiderivative. Therefore, the antiderivative of \( e^x \), denoted as \( F(x) \), is simply \( F(x) = e^x + C \), where \( C \) is the constant of integration.
03

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \) over an interval \( [a, b] \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).
04

Evaluate the Integral

To evaluate \( \int_{0}^{1} e^x \, dx \), substitute the upper limit and lower limit into the antiderivative: \( F(1) - F(0) = e^{1} - e^{0} = e - 1 \).

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Key Concepts

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

Understanding the Antiderivative
An antiderivative is a fundamental concept in calculus that helps us reverse the process of differentiation. When we differentiate a function, we find how it changes at any point. But when we find an antiderivative, we essentially look for a function whose derivative gives us the original function. In the given exercise, the function we need to handle is the exponential function, specifically, the integrand is the function \( f(x) = e^x \). Thanks to the unique characteristic of the exponential function \( e^x \), its derivative is the same as the function itself. Thus, the antiderivative of \( e^x \), often denoted as \( F(x) \), will also be \( e^x + C \), where \( C \) represents the constant of integration. The constant of integration appears because the process of differentiation eliminates constants, so when reversing the process, they reappear as an unknown constant.
The Fundamental Theorem of Calculus Explained
The Fundamental Theorem of Calculus is a bridge between differential and integral calculus. It tells us that differentiation and integration are inverse processes. This theorem comes in two parts, but for definite integrals, the second part is more important. It states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \([a, b]\), then the definite integral of \( f(x) \) from \( a \) to \( b \) is given by the difference \( F(b) - F(a) \). This is useful because it transforms a potentially complicated calculation of total area under the curve \( f(x) \) into a simple subtraction once we find an antiderivative. In our exercise, once we find the antiderivative of \( e^x \) as \( e^x + C \), we apply the Fundamental Theorem by calculating \( e^1 - e^0 \). This yields \( e - 1 \), which means the area under the curve \( e^x \) from \( x = 0 \) to \( x = 1 \) is exactly \( e - 1 \).
Characteristics of the Exponential Function
Exponential functions are extremely important in mathematics due to their unique properties and applications in various fields. A key property of the exponential function \( e^x \) is that it is its own derivative and antiderivative, making calculations involving derivatives and integrals straightforward. The exponential function grows very rapidly; when you graph \( e^x \), you'll notice it starts slowly but then increases at an accelerating rate. This characteristic is crucial in modeling a wide range of exponential growth scenarios, from population growth to radioactive decay. In our exercise, this property simplifies finding the antiderivative. For other functions, determining the antiderivative might require more complex techniques or integration rules. But with \( e^x \), you know right away that its antiderivative \( F(x) \) is \( e^x + C \). This simplicity is a reason why exponential functions are emphasized heavily in calculus and why they are often used for introductory concepts in integration.

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Most popular questions from this chapter

In a particular regional climate, the temperature varies between \(-22^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), averaging \(\mu=11^{\circ} \mathrm{C}\). The number of days \(F(T)\) in the year on which the temperature remains below \(T\) degrees centigrade is given (approximately) by $$F(T)=\int_{-22}^{T} f(x) d x \quad(-22 \leq T \leq 40)$$ where $$f(x)=12.72 \exp \left(-\frac{(x-11)^{2}}{266.4}\right) $$ Notice that \(F(T)\) is the sort of area integral that we studied in Section 5.4 . a. Use Simpson's Rule with \(N=20\) to approximate \(F(40) .\) What should the exact value of \(F(40)\) be? b. Heat alerts are issued when the daily high temperature is \(36^{\circ} \mathrm{C}\) or more. On about how many days a year are heat alerts issued? c. Suppose that global warming raises the average temperature by \(1^{\circ} \mathrm{C}\), shifting the graph of \(f\) by 1 unit to the right. The new model may be obtained by simply replacing \(\mu\) with 12 and using [-21,41] as the domain (see Figure 13). What is the percentage increase in heat alerts that will result from this \(1^{\circ} \mathrm{C}\) shift in temperature?

The integral \(\int_{a}^{b}\left(f_{1}(x)-f_{2}(x)\right) d x\) represents the area of a region in the \(x y\) -plane that is bounded by the graphs of \(f_{1}\) and \(f_{2}\). Express the area of the region as an integral of the form \(\int_{c}^{d}\left(g_{1}(y)-g_{2}(y)\right) d y .\) For example, the integral \(\int_{0}^{1}\left(x-x^{2}\right) d x\) represents the area of the shaded region in Figure \(11 .\) This area can also be represented as \(\int_{0}^{1}(\sqrt{y}-y) d y.\) $$ \int_{-1}^{1}(e-\exp (|x|)) d x $$

Speed measurements of a runner taken at half-second intervals during the first 5 seconds of a sprint are provided in the following table.$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text { Time (s) } & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 & 4.0 & 4.5 & 5.0 \\\\\hline \text { Speed (m/s) } & 0 & 5.26 & 6.67 & 7.41 & 8.33 & 8.33 & 9.52 & 9.52 & 10.64 & 10.64 & 10.87 \\\\\hline\end{array}$$ About how many meters did the athlete run during that 5-second interval? Use Simpson's Rule.

Treat the \(y\) variable as the independent variable and the \(x\) variable as the dependent variable. By integrating with respect to \(y,\) calculate the area of the region that is described. The region between the curves \(x=\cos (y)\) and \(x=\sin (y)\) \(\pi / 4 \leq y \leq 5 \pi / 4\)

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 between the graph of \(f\) and the \(x\) -axis. \(f(x)=x /\left(x^{2}+1\right)^{2} \quad I=[-1,3]\)

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