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Multiple Choice
For which values of p does the improper integral converge?
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Verified step by step guidance
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Step 1: Recognize that the integral \( \int_1^{\infty} x^{3p+1} \, dx \) is improper because the upper limit of integration is infinity. To determine convergence, analyze the behavior of the integrand as \( x \to \infty \).
Step 2: Rewrite the integral in terms of the general formula for integration of power functions. For \( \int x^n \, dx \), the result is \( \frac{x^{n+1}}{n+1} \) provided \( n \neq -1 \). Here, \( n = 3p+1 \).
Step 3: Apply the formula for the definite integral. Substitute the limits of integration: \( \int_1^{\infty} x^{3p+1} \, dx = \lim_{b \to \infty} \left[ \frac{x^{3p+2}}{3p+2} \right]_1^b \).
Step 4: Analyze the term \( \frac{x^{3p+2}}{3p+2} \) as \( x \to \infty \). For the integral to converge, \( x^{3p+2} \) must approach 0 as \( x \to \infty \). This happens if \( 3p+2 < 0 \), which simplifies to \( p < -\frac{2}{3} \).
Step 5: Verify the condition \( p > -1 \) given in the problem. Since \( p < -\frac{2}{3} \) is stricter than \( p > -1 \), the convergence condition is \( p < -\frac{2}{3} \).