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The series is convergent. (A). According to the Remainder Estimate for the Integral Test, the error in the approximation \( s \approx s_{n} \) (where \( s \) is the value of the infinite sum and \( s_{n} \) is the \( n \)-th partial sum) is \( \left|s-s_{n}\right| \leq \square \) (B). Find the smallest value of \( n \) such that this upper bound is less than 0.001 . \( n=\square \)

Ask by Chambers Gardner. in the United States
Mar 07,2025

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**(A)** The error in the approximation \( s \approx s_{n} \) is bounded by \( \frac{1}{n} \). **(B)** The smallest value of \( n \) such that the error is less than 0.001 is \( n = 1001 \).

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Bonus Knowledge

To estimate the error in the approximation of a convergent series using the Remainder Estimate for the Integral Test, we can express the error as \( \left|s-s_{n}\right| \leq \int_{n}^{\infty} f(x) \, dx \), where \( f(x) \) is the function corresponding to the series' terms. By effectively plotting the function and calculating the integral, we can narrow down how close \( s \) is to \( s_n \). To find the smallest \( n \) such that this upper bound is less than 0.001, proceed by evaluating the integral from \( n \) to \( \infty \) and solving for \( n \). This typically involves finding a value for which \( \int_{n}^{\infty} f(x) \, dx < 0.001 \). Start with trial values for \( n \) and iterate until you meet the condition.

preguntas relacionadas

Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
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