Final answer:
To determine the convergence or divergence of the integral ∫₀⁴ 1/(x²-x-2) dx, we can start by factoring the denominator. The integral converges on the interval [0, 4], except at x = -1.
Step-by-step explanation:
To determine the convergence or divergence of the integral ∫₀⁴ 1/(x²-x-2) dx, we can start by factoring the denominator. The denominator factors into (x+1)(x-2). Next, we need to determine if the integral converges or diverges at the endpoints of the interval [0, 4], by checking if the function has any vertical asymptotes or if the integrand becomes unbounded at those points.
For x = -1, we see that the function becomes unbounded, so the integral diverges at x = -1. For x = 2, the function is bounded and does not have any vertical asymptotes, so the integral converges at x = 2.
Therefore, the integral ∫₀⁴ 1/(x²-x-2) dx converges on the interval [0, 4], except at x= -1.