Question

i need help on 29,31,35,37

11 27. wate the following integra ntegrals using 29-62. Definite integrals Evaluate the Fundamental Theorem of Calculus. 30. Jo (3x² + 2x) dx Ji (29.) [ 4x ? dx Jo 30. (°(3x² + 2x) dx 31, L* 8x1/30 32 (19 -514 2x 33. Jo (x + Vrsar 34. / A 2 cost ap 2 de 36. (2+ + de 37. [ (x2 - 444 -2 38. Sinsear 30. I 2 (13 – 8) di 40. Salt-2) 11- V VX JO

Answer 1

Here we use the fundamental theorem of calculus to evaluate the definite integral.
• interval of fox) Fundamental theorem of calculus : - If f(x) is continuons over the [a, b ] and F(x) is the antiderivative then the theorem says that 1 feas dx = F (6) - F (0) 3 + 1²403 da det f(a) = 4x3, Continuous on (0,2] the antiderivative of f(n) be F(2) is given by F(x) = J 4x² dx [ Power rule 7 = (4) () te landa = 2t! te = ate By Fundamental theorem of calculus 423 dr = F(2) – F10) [c is the 7 = [(2) 4+c]- [cojtach arbitrary Constant 5 16+6-0-C = 16 . n -1 naam 18 4x² dm = 16 si het f(n) = 803 is continuons on [18] Anti derivative F(A) is given by F(n) - 5 8x43 dx B C in the arbitrary Constant. = (8).[**] + = (8) (24) alte = 68% te

By Fundamental theorem of Calculus S® 8x>de = F(8) - F(1) = [() (8) +e] - [6) (w)*+c] = 6) (23)+6-6- = (6) (24) -6 = (6).(16-1) = (6)(15) = 90 8 x 13 dx = go Let f(x) = 2 is continuous on [19] Antiderivative, F(a) is given by - F(A) = 2 da e is the arbitrary = (2) x Y2 dx Constant- = (2) [a2+ Tre 3 (2) ( ) +C = (2)() va + - 45ate By Fundamental theorem of caseins So fare dx = F(9) - FCI) = [4 55+c] - [4 T +c] = (4) (3) +(-4- - 12-4 = 8 Sip Rada 28 UY

39 S (a?4) dre Constant. Let f(x) = x²-4 is continuous on [-2, 2] Antiderivative of f(x) is given by F() = 5 (?-4) dx 1 = (x²da-45 da (is the arbitrary = -4x)+C By Fundamental theorem of Calculus 5 (224) dx = F(2) – F(-2) = [03*<(4)(2)+c]-[632-4(-2)+c] = ( -8 +5)-(-+8+c) = 8 + C + D -8-2 - 1 - 16 = 167.98 = - 32 $ (24) da = – 30