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Chapter 7: Problem 5

Evaluate the following integrals or state that they diverge. $$\int_{1}^{\infty} x^{-2} d x$$

Short Answer

Expert verified
Question: Evaluate the improper integral or state that it diverges: $$\int_{1}^{\infty} x^{-2} dx$$ Answer: The integral converges and is equal to \(1\): $$\int_{1}^{\infty} x^{-2} dx = 1$$

Step by step solution

01

Rewrite the given integral

We are given the following integral: $$\int_{1}^{\infty} x^{-2} dx$$ We can rewrite it as: $$\int_{1}^{\infty} \frac{1}{x^2} dx$$
02

Integrate the function

Now, let's integrate the function: $$\int \frac{1}{x^2} dx = \int x^{-2} dx$$ Using the power rule for integration (\(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)) for \(n \neq -1\), we have: $$\frac{x^{-2+1}}{-2+1} = -\frac{x^{-1}}{1} = -x^{-1}$$ So, the indefinite integral is: $$\int x^{-2} dx = -x^{-1} + C = -\frac{1}{x} + C$$
03

Apply the limits of integration

Now that we have found the indefinite integral, let's apply the limits of integration using the Fundamental Theorem of Calculus: $$\int_{1}^{\infty} \frac{1}{x^2} dx = -\frac{1}{x}\Big|_1^\infty$$
04

Evaluate the limit

To evaluate the definite integral, we need to evaluate the limit as the upper bound approaches infinity: $$\lim_{t\to\infty} \left[ -\frac{1}{x}\right] \Big|_1^t = \lim_{t\to\infty} \left( -\frac{1}{t} - \left(-\frac{1}{1}\right)\right)$$ Simplify the expression: $$\lim_{t\to\infty} (1-\frac{1}{t})$$ As \(t\) approaches infinity, the second term \(\frac{1}{t}\) approaches zero: $$\lim_{t\to\infty} (1-\frac{1}{t}) = 1$$
05

Conclude the result

Since the limit exists and is equal to \(1\), the integral converges: $$\int_{1}^{\infty} x^{-2} dx = \int_{1}^{\infty} \frac{1}{x^2} dx = 1$$

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Most popular questions from this chapter

Circumference of a circle Use calculus to find the circumference of a circle with radius \(a.\)

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

Exact Simpson's Rule Prove that Simpson's Rule is exact (no error) when approximating the definite integral of a linear function and a quadratic function.

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \cos x \cos 2 x d x$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u\). Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$
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