Question

Let f(x) be a function that is defined and has a continuous derivative on the interval (2,[infinity]).

Assume also that f(3) = -6 |f(x)| < x⁶ + 2 and ∫₃[infinity] f(x)e⁻ˣ/³ dx = -8
Determine the value of ∫₃[infinity] f'(x)e⁻ˣ/³ dx =

Answer 1

To calculate the value of ∫₃ ∞ f'(x)e⁻ˣ/³ dx, we can use integration by parts. Given that ∫₃ ∞ f(x)e⁻ˣ/³ dx = -8, the value of ∫₃ ∞ f'(x)e⁻ˣ/³ dx is -8.

Step-by-step explanation:

To calculate the value of ∫₃ ∞ f'(x)e⁻ˣ/³ dx, we can use integration by parts. Let u = f'(x) and dv = e⁻ˣ/³ dx. Then du = f''(x) dx and v = -3e⁻ˣ/³. Applying the formula for integration by parts, we have: ∫₃ ∞ f'(x)e⁻ˣ/³ dx = [-3f'(x)e⁻ˣ/³]₃ ∞ - ∫₃ ∞ (-3)e⁻ˣ/³ f''(x) dx. Since f(x) has a continuous derivative on the interval (2, ∞), we can evaluate the integral by taking the limit as the lower bound approaches infinity. The first term on the right side of the equation becomes 0, and the integral simplifies to ∫₃ ∞ (-3)e⁻ˣ/³ f''(x) dx. Given that ∫₃ ∞ f(x)e⁻ˣ/³ dx = -8, we can equate this to ∫₃ ∞ (-3)e⁻ˣ/³ f''(x) dx and solve for ∫₃ ∞ f'(x)e⁻ˣ/³ dx: -8 = ∫₃ ∞ (-3)e⁻ˣ/³ f''(x) dx. Therefore, the value of ∫₃ ∞ f'(x)e⁻ˣ/³ dx is -8.