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Chapter 1: Problem 20

Use the Mean-Value Theorem for Integrals to prove that $$ \int_{0}^{\pi / 2} e^{x} \cos x d x=e $$ for some \(y\) in \((0, \pi / 2)\).

Short Answer

Expert verified
The equation \( e^y \cos y = \frac{2e}{\pi} \) confirms \( y \) exists in \((0, \pi/2)\).

Step by step solution

01

Understanding the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), then there exists at least one point \( c \) in \((a, b)\) such that \[ \int_{a}^{b} f(x) \; dx = f(c)(b - a). \] Our job is to find \( c \) such that this holds for the given integral.
02

Identify the Function and Interval

In the exercise, the function we are dealing with is \( f(x) = e^x \cos x \) and the interval is \([0, \pi/2]\). The theorem tells us there is some point \( y \) in \((0, \pi/2)\) for which \( \int_0^{\pi/2} e^x \cos x \; dx = e^y \cos y \cdot \frac{\pi}{2} \).
03

Evaluate the Definite Integral

We are given that \( \int_{0}^{\pi/2} e^x \cos x \; dx = e \). Hence, this means that there's a \( y \) such that \( e^y \cos y \cdot \frac{\pi}{2} = e \).
04

Solving for \( y \)

Rearrange the equation \( e^y \cos y \cdot \frac{\pi}{2} = e \) to isolate \( e^y \cos y \): \[ e^y \cos y = \frac{2e}{\pi}. \] This equation guarantees there exists some \( y \) in \((0, \pi/2)\) satisfying this identity, according to the theorem.

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Key Concepts

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

Continuous Function
A continuous function is a fundamental concept in calculus. It refers to a type of function where small changes in the input result in small changes in the output.
  • This means that the graph of a continuous function has no breaks, jumps, or holes.
  • In mathematical terms, a function is continuous at a point if the limit as you approach the point is equal to the function鈥檚 value at that point.
In the context of the Mean-Value Theorem for Integrals, continuity is crucial. The theorem requires the function to be continuous over the interval \([a, b]\).
This ensures the integral \(\int_a^b f(x) \, dx\) is well-defined and the theorem can be applied.
In the given exercise, \(f(x) = e^x \cos x\) is continuous over \[0, \pi/2\]. This allows us to confidently apply the theorem and find the value of \(y\) where the integral equals a function evaluation.
Definite Integral
A definite integral is a concept in calculus that calculates the accumulation of quantities.
  • It is represented as the area under the curve of a function between two points on the x-axis.
  • The notation for a definite integral is \(\int_a^b f(x) \, dx\), where \(a\) and \(b\) are the bounds of the interval.
In the exercise, the definite integral involved is \(\int_0^{\pi/2} e^x \cos x \, dx\).
This integral is crucial because the Mean-Value Theorem for Integrals relates the value of this integral to the function evaluation at a specific point within the interval.
The given integral equals \(e\), meaning there is some value \(y\) within \(0, \pi/2\) such that the function evaluated at \(y\) times the length of the interval equals this value.
Function Evaluation
Function evaluation is the process of determining the value of a function for a specific input. In the context of the Mean-Value Theorem for Integrals, it is a critical step.
  • The theorem asserts there exists a point \(c\) in the interval \(a, b\) where the integral can be expressed as a product of the function evaluated at \(c\) and the interval length.
  • This helps to find the exact value of a continuous function at some point within the bounds.
In the exercise, we are tasked with evaluating \(y\) such that \(e^y \cos y \cdot \frac{\pi}{2} = e\).
This requires solving for \(y\) which makes the function \(e^y \cos y\) equal to \(\frac{2e}{\pi}\).
The theorem guarantees this \(y\) exists, providing a powerful method to connect the definite integral to specific function values within the interval.

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Most popular questions from this chapter

Use Taylor's Theorem with \(n=2\) to prove that the inequality \(1+x

Let a sequence \(x_{n}\) be defined inductively by \(x_{n+1}=F\left(x_{n}\right) .\) Suppose that \(x_{n} \rightarrow x\) as \(n \rightarrow \infty\) and \(F^{\prime}(x)=0 .\) Show that $$ x_{n+2}-x_{n+1}=\mathcal{O}\left(x_{n+1}-x_{n}\right) $$ Hint: Use the Mean-Value Theorem and assume that \(F\) is a continuously differentiable function.

Define a sequence inductively by the equation \(x_{n+1}=x_{n}+x_{n}^{-1}\), where \(x_{0}>0\). Determine the behavior of \(x_{n}\) as \(n \rightarrow \infty\).

Show that \(\lim _{x \rightarrow 1}(4 x+2)=6\) by means of an \(\varepsilon-\delta\) proof.

In solving differential equations, one often obtains solutions in implicit form. Show that the equation $$ 2 x^{3} y^{2}+x^{2} y+e^{x}=c $$ defines a solution of the differential equation $$ \frac{d y}{d x}=-\left(6 x^{2} y^{2}+2 x y+e^{x}\right) /\left(4 x^{3} y+x^{2}\right) $$
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