How to solve these 2 different question?

(3 marks) Consider the improper integral OO I = 8 f(x) dac where f(2) = 12 x2 -36 i) First let us notice that for x 2 8 , f (x) takes positive values only + allows us to compare f (a) to g(ac) = f (a) Choosing P = xp . so that lim = 12. OO By considering the p-integral integral M = g(2) dac, the limit form of the Comparison test states that both I and M converge. ii) Note that f (20) = 7 1 x - 6 x +6 Now let c > 8. Then I(c) = f(x) dx = [In |F(x)|18 = In G(c) where F (2) = and G (c) = By letting c - oo we can get that OO I = 8 f(x) dx = (for co please write infinity).(2 marks) Integration by parts can be used to evaluate the integral In(1 + 4x) da . Indeed, integration by parts gives (simplify your answer where possible) In(1 + 4x) dx = dac and hence In(1 + 4x) dr = GSP+ C