5 e-3/+3 dt 14 a) .5 e-3/+3 e-3/+3 dt = lim T-aJb dt , 14 where a = 0 b = 0 and C 5 FORMATTING: To enter a one-sided limit value such as 17 or 17, write 1^+ or 1^- in Mobius. b) To compute the integral e-3/+3 .it 14 that you have found in (a), we need to use the change of variable u = -3/t^3 c) With the change of variable that you have found in (b), we have ce-3/+3 b -dt = 14 B f(u)du where B = C = -3/125 and f(u) = 1/9*e^u d) With the information found in (c), we find that ce-3/13 +4 -dt = B f(u)du = 1/9/e^(3/125) e) Hence, e-3/13 lim dt = lim T-ajb 14 T-aJB f(u)du = 1/9*e^(3/125) f) In conclusion, 5 e-3/13 o dt = 1/9*e^(3/125) 14 Write Diverges if the integral is divergent