Question

there are two parts please helpe me with both the parts and request you to write neatly

Answer 1

(a) Consider the integral integral 3 squareroot x e^3 squareroot x dx Substituting 3 squareroot x = t i.e. x = t^3 1/3 x^- 2/3 dx = dt 1/3 x^2/3 dx = dt dx = 3 x^2/3 dt dx = 3 (t^3)^2/3 dt dx = 3 t^2 dt So, the integral becomes, integral t e^t middot (3 t^2 dt = 3 integral t^3 e^t dt = 3 [(t^3) (e^t) - (3 t^2) (e^t) + (6 t) (e^t) - (6) (e^t)] = 3 [(t^3 - 3 t^2 + 6 t - 6) e^t] = 3 (t^3 - 3 t^2 + 6 t - 6) e^t Put value of t we get, \r\n integral 3 squareroot x e^3 squareroot x dx = 3 (3 squareroot x)^3 - 3 (3 squareroot x)^2 + 6 (3 squareroot x) - 6) e^3 squareroot x Now taking limit from 0 to 8. integral^8_0 3 squareroot x e^3 squareroot x dx = [3 ((3 squareroot x)^3