Question

Write an inequlity comparing n²-6/n⁵+6 with 1/n³ for n ≥ 1. (Express numbers in exact form. Use symbolic notation and fractions where needed.) Use this inequality to determine whether the integral ₆∫⁸ e⁻⁽ˣ ⁺ ¹/ˣ⁾dx converges.

Answer 1

Comparing ² - 6/µ + 6 with 1/³ for ≥ 1 shows that the former is less than the latter. However, this comparison does not provide a definitive answer as to whether the integral 6∫⁸ e⁻(x + 1/x)dx converges; further analysis is required.

Step-by-step explanation:

To compare n² - 6/n⁵ + 6 with 1/n³ for n ≥ 1, we can begin by simplifying the expressions. Since n is at least 1, n² and n⁵ will always be positive, so we can ignore the addition and subtraction of 6 when comparing magnitudes:

² - 6 < ² < ² + 6

⁵ - 6 < ⁵ < ⁵ + 6

We can see that as n increases, n² - 6/µ + 6 becomes much smaller than 1/³ due to the higher power of n in the denominator. Therefore, for all n ≥ 1:

n² - 6/n⁵ + 6 < 1/n³

Now, consider the integral 6∫⁸ e⁻(x + 1/x)dx. If the integrand e⁻(x + 1/x) is greater than a function whose integral we know converges, then we can say that our original integral also converges. However, by comparing with 1/x³, which diverges as x → ∞, we don't have enough information to determine the convergence of our integral directly, since 1/x³ is not guaranteed to dominate e⁻(x + 1/x) over the interval [6, 8]. For a definite answer, further analysis using comparison tests or direct integration would be necessary.