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Chapter 8: Problem 1

Find the average of the function \(f:[1,2] \rightarrow \mathbb{R}\) defined by \(f(x):=1 / x\).

Short Answer

Expert verified
The average value of the function \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\) is \(\ln(2)\).

Step by step solution

01

Integrate the function over the given interval

To integrate the function \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\), we need to find the definite integral: \[\int_{1}^{2} \frac{1}{x} dx\]
02

Evaluate the integral

We can solve the integral using the natural logarithm. The antiderivative of \(\frac{1}{x}\) is \(\ln(x)\). Now we can evaluate the definite integral: \[\int_{1}^{2} \frac{1}{x} dx = \ln(x)\Big|_1^2 = \ln(2) - \ln(1)\] Since \(\ln(1) = 0\), the integral evaluates to: \[\int_{1}^{2} \frac{1}{x} dx = \ln(2)\]
03

Divide by the length of the interval

The length of the interval is \(2 - 1 = 1\). To find the average value of the function, we divide the result of the integral by the length of the interval: \[\frac{\int_{1}^{2} \frac{1}{x} dx}{(2 - 1)} = \frac{\ln(2)}{1}\]
04

Simplify the expression

Since the length of the interval is 1, the average value of the function remains the same: Average of \(f(x) = \frac{1}{x}\) over \([1, 2]\) = \(\ln(2)\)

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Most popular questions from this chapter

Let \(a \in \mathbb{R}\) with \(a>0\). Find the centroid of the region bounded by the curves given by \(y=-a, x=a, x=-a\), and \(y=\sqrt{a^{2}-x^{2}}\).

A solid body lies between the planes given by \(y=-2\) and \(y=2\). Each of its slices by a plane perpendicular to the \(y\) -axis is a disk with a diameter extending between the curves given by \(x=y^{2}\) and \(x=8-y^{2}\). Find the volume of the solid body.

Find the arc length of each of the curves mentioned below. (i) the cuspidal cubic given by \(y^{2}=x^{3}\) between the points \((0,0)\) and \((4,8)\) (ii) the cycloid given by \(x=t-\sin t, y=1-\cos t,-\pi \leq t \leq \pi\) (iii) the curve given by \((y+1)^{2}=4 x^{3}, 0 \leq x \leq 1\), (iv) the curve given by \(y=\int_{0}^{x} \sqrt{\cos 2 t} d t, 0 \leq x \leq \pi / 4\).

If \(f, g:[a, b] \rightarrow \mathbb{R}\) are integrable functions, then show that \(\operatorname{Av}(f+g)=\) \(\operatorname{Av}(f)+\operatorname{Av}(g)\), but \(\operatorname{Av}(f g)\) may not be equal to \(\operatorname{Av}(f) \operatorname{Av}(g)\)

If a piecewise smooth curve \(C\) is given by \(r=p(\theta), \theta \in[\alpha, \beta]\), and \(\ell(C)=\) \(\int_{\alpha}^{\beta} \sqrt{p(\theta)^{2}+p^{\prime}(\theta)^{2}} d \theta \neq 0\), then show that the centroid \((\bar{x}, \bar{y})\) of \(C\) is given by $$ \bar{x}=\frac{1}{\ell(C)} \int_{\alpha}^{\beta} p(\theta) \cos \theta \sqrt{p(\theta)^{2}+p^{\prime}(\theta)^{2}} d \theta $$ and $$ \bar{y}=\frac{1}{\ell(C)} \int_{\alpha}^{\beta} p(\theta) \sin \theta \sqrt{p(\theta)^{2}+p^{\prime}(\theta)^{2}} d \theta $$
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