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Chapter 7: Problem 47

Evaluate. \(\int_{0}^{1} \frac{e^{x}+1}{e^{x}} d x\).

Short Answer

Expert verified
The definite integral of the function \(\frac{e^x + 1}{e^x}\) on the interval \([0, 1]\) can be computed as follows: Simplify the integrand to \(1 + e^{-x}\), integrate with respect to x to obtain the antiderivative \(x - e^{-x} + C\), and then evaluate the difference between the antiderivative at the upper (\(x = 1\)) and lower (\(x = 0\)) bounds. This results in a final value of \[2 - e^{-1}\] for the definite integral.

Step by step solution

01

Simplify the function to integrate

To simplify the integrand function, we can rewrite it as: \[\frac{e^x + 1}{e^x} = \frac{e^x}{e^x} + \frac{1}{e^x} = 1 + e^{-x}\]
02

Integrate the simplified function

Now, we will integrate \(1 + e^{-x}\) with respect to x: \[\int(1 + e^{-x}) dx\] The integration of \(1\) with respect to \(x\) is: \[x\] The integration of \(e^{-x}\) with respect to \(x\) is: \[-e^{-x}\] Combine both results: \[F(x) = x - e^{-x} + C\] Where \(F(x)\) is the antiderivative and \(C\) is a constant of integration. In this case, we don't need to find the value of \(C\) as we will be taking the difference between two values of the antiderivative, which will cancel out the constant.
03

Evaluate the antiderivative at the upper and lower bounds

Now, we will find the difference between the antiderivative at the upper bound (1) and the lower bound (0). Evaluate at the upper bound, \(x = 1\): \[F(1) = 1 - e^{-1}\] Evaluate at the lower bound, \(x = 0\): \[F(0) = 0 - e^{0} = -1\]
04

Find the difference between the antiderivative at the upper and lower bounds and conclude

Take the difference between the antiderivative at the upper and lower bounds to find the definite integral: \[\int_{0}^{1} \frac{e^x + 1}{e^x} dx = F(1) - F(0) = (1 - e^{-1}) - (-1) = 2 - e^{-1}\] Thus, the definite integral of the given function on the interval \([0, 1]\) is: \[2 - e^{-1}\]

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Key Concepts

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

Definite Integrals
A definite integral finds the area under a curve between two specific points on the x-axis. It is denoted by the integral sign with upper and lower limits, like \[ \int_{a}^{b} f(x) \, dx \]where \(a\) and \(b\) are the bounds. In our exercise, these bounds are 0 and 1. The result is a concrete value representing this area.

Definite integrals also have crucial properties, such as:
  • The integral from \(a\) to \(b\) can be viewed as \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)
  • If \(f(x)\) is continuous on \([a, b]\), then the definite integral exists
In essence, by evaluating the antiderivative at these points, we obtain the precise area under the curve, rectangle, or any path defined by the function.
Antiderivative
An antiderivative is a function whose derivative yields the original function. It is the reverse process of differentiation. Symbolically, if \(F(x)\) is the antiderivative of \(f(x)\), then \(\frac{d}{dx}[F(x)] = f(x)\).

When finding definite integrals, the antiderivative plays a key role. For the given function \(1 + e^{-x}\), we found:
  • The antiderivative of \(1\) is simply \(x\).
  • The antiderivative of \(e^{-x}\) is \(-e^{-x}\).
Hence, the antiderivative \(F(x)\) is \(x - e^{-x} + C\), where \(C\) is a constant. However, in definite integrals, this constant cancels out when subtracting \(F(b) - F(a)\). This simplification helps in finding exact values efficiently.
Exponential Functions
Exponential functions involve the constant \(e\), where \(e\) is approximately 2.718, and it forms the base of natural logarithms. These functions have unique properties, including rapid growth and decay, which play a fundamental role in calculus.

Here's why exponential functions are important:
  • They simplify differentiation and integration. For example, \(\frac{d}{dx}[e^x] = e^x\) and \(\int e^x \, dx = e^x + C\).
  • In our problem, instead of dealing with complex expressions, we easily broke down \(\frac{e^x + 1}{e^x}\) into \(1 + e^{-x}\), a much simpler form.
This manipulation allows easier integration. The exponential decay, \(e^{-x}\), smoothly handles rate changes, making calculus operations more straightforward.

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Most popular questions from this chapter

Determine the following: (i) the domain; (ii) the intervals on which \(f\) increases, decreases; (iii) the extreme values; (iv) the concavity of the graph and the points of inflection. Then sketch the graph, indicating all asymptotes. \(f(x)=x^{2} e^{-x}\).

Show that the average slope of the logarithm curve from \(x \quad a\) to \(x=b\) is $$\frac{1}{b-a} \ln \left(\frac{b}{a}\right)$$

Sketch the region bounded above by \(y=8 /\left(x^{2}+4\right)\) and bounded below by \(4 y=x^{2} .\) What is the area of this region?

Let \(\Omega\) be the region below the graph of \(y=e^{x}\) from \(x=0\) 10 \(x=1\) (a) Find the volume of the solid generated by revolving \(1 \mathrm{g} \Omega\) about the \(x\) -axis. (b) Set up the definite integral that gives the volume of the solid generated by revolving \(\Omega\) about the \(y\) -axis using the shell method. (You will see how to evaluate this integral in Section \(8.2 .)\)

Find \(f^{-1}\). $$f(x)=\frac{3 x}{2 x+5}, \quad x \neq-5 / 2$$
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