Final answer:
a) The integral ∫₀π/² tan(x)dx is improper and divergent. b) The integral ∫₀³9−x²x dx is improper and divergent. c) The integral ∫−∞∞(x²+1)(x²+4)x² dx is improper and convergent.
Step-by-step explanation:
a) The integral ∫₀π/² tan(x)dx is improper because the function tan(x) is not defined at x = π/2. As x approaches π/2, the value of tan(x) goes to infinity. To evaluate the convergence or divergence of the integral, we need to check if the function tan(x) has a vertical asymptote at x = π/2. Since tan(x) is unbounded as x approaches π/2, the integral diverges. Therefore, the integral is divergent.
b) The integral ∫₀³9−x²x dx is improper because the function has a vertical asymptote at x = 3. The denominator x in the integrand approaches 0 as x approaches 3. To evaluate its convergence or divergence, we need to check if the function has a vertical asymptote at x = 3. Since the denominator approaches 0, the integral diverges. Therefore, the integral is divergent.
c) The integral ∫−∞∞(x²+1)(x²+4)x² dx is improper because the limits of integration are from negative infinity to positive infinity. To evaluate its convergence or divergence, we start by expanding the integrand. (x²+1)(x²+4)x² = x^2(x^2+1)(x^2+4) = x^6 + 5x^4 + 4x^2. Since the degree of the highest power of x in the integrand is even, we can split the integral into two parts: ∫−∞∞ x^6 dx, ∫−∞∞ 5x^4 dx, and ∫−∞∞ 4x^2 dx. Each of these integrals converges. Therefore, the integral converges and its exact value will depend on the evaluation of each of the individual integrals.