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Chapter 7: Problem 4

For what values of \(p\) does \(\int_{1}^{\infty} x^{-p} d x\) converge?

Short Answer

Expert verified
The integral converges for \(p > 1\).

Step by step solution

01

Find an antiderivative

We need to find an antiderivative for \(x^{-p}\) in order to evaluate the integral. The antiderivative of \(x^{-p}\) is \(\frac{x^{-p+1}}{-p+1}\), which we can verify by taking the derivative, assuming \(p \neq 1\).
02

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we can now evaluate the integral: $$\int_{1}^{\infty} x^{-p} d x = \lim_{b \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_1^b,$$ where we apply the limit since it is an improper integral.
03

Evaluate the limit

Taking the limit as b approaches infinity, we get: $$\lim_{b \to \infty} \left[ \frac{b^{-p+1}}{-p+1} - \frac{1^{-p+1}}{-p+1} \right].$$
04

Determine when the limit converges

To make the expression converge, we want the first term of the expression to approach 0: $$\lim_{b \to \infty} \frac{b^{-p+1}}{-p+1}.$$ Notice that this limit converges to 0 if \(-p+1 < 0\) or \(p > 1\). Therefore, if \(p > 1\), the integral converges and diverges otherwise. So, the integral \(\int_{1}^{\infty} x^{-p} d x\) converges for \(p > 1\).

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Most popular questions from this chapter

Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

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