Please help me answer the following questions:
Starting from 2- to 8 (even numbers ), 48, 50, 52, 56, 58, 60, 70, 72, 74
I'm going to be attaching some similar examples from the same page. It would be really helpful if you followed the same format.
Find the area under the given curve over the indicated interval. Evaluate. Then interpret the results. 1. y = 4; [1, 3] 2. y = 5; [1, 3] . ( x - x2 ) dx 16 . [ ( x 2 - x ) dx 0.5- _ y = x -x2 y=x-x 0.5 0.5 15 x 3. y = 2x; [1, 3] 4. y = x2; [0, 3] 70.5 - y = x2 y' = 2x/ -3 Ti/2 TITTTTT 17. Jo cos x dx ITTIT 1 2 5. y = x2; [0, 5] 6. y = x3; [0, 2] y = cos x 0.5 30 y = x2 30 - y =x3 25 20 - 20 - 15 -0.5- 10 10- 5 - -1/ -2-1 1 2 3 4 5 x 1 2 7. y = x'; [0, 1] 8. y = 1 - x2 [-1, 1] YA yEx' 18. - 2e3x dx 1.0 y= 1-x2 0.51 2 x -05 E 0.5 1.0 x -2 9. y = 4 - x2; [-2,2] 10. y = ex; [0, 2] 11. y = ex; [o, 3] 12. y = ~; [1, 4]Find the area under the graph over the indicated interval. 47. y = x3; [0, 2] 48. y = x4; [0, 1] 49, y = x2 + x + 1; [2, 3] 50. y = 2- x - x2; [-2, 1] 51. y = 5 - x2; [-1, 2] 52. y = ex; [-2, 3] 53. y = ex; [-1, 5] 54. y = 2x + -2; [1, 4] Evaluate. Some algebra may be required before finding the integral. - X 55. dx 56. dx 12 x- 1 -16 57. (x - 1) V x dx 58. (x+ 2)3 dx 81 x2 - 1 x3+ 8 59. dx 60. dx V x Jo x+2\fPg # 369 1/ 1 1 / 4 = 4 ; [ 1 1 3] 3 ) y = 2 x ] [ 1, 3] 1) 3 3 A = b - h = z ( 41 ) = 8 A = ( 2 xd x = z x2 Area = (udx - 186) 3 D = 13 = X2 = UX = 4(3 ) - 4(1) = 18 = (3)2 - (1)2 = 8 5 ) y = x 2 ; Co,s] 7 1 = x 3 : [o, 1] 25- 1=x ? 10 0.5 -1 12 3 45 - 0.5 F 05 3 10 A= S x2dx= x21. X 315 A = S x 3 dx = * 4!' 2 +1 3 = [15)3] -[ 1023 ] = 1 125 3\f2 3 2 1-1- 2.25 + 3- 375 +5 -3= 3 3 17) 317 / 2 cosxdx ACOSX 0.5 X 0.5+ 2 J-COSX A = ( 1/ 2 Losx + ( 31 / 2 Cosxax A = Sinx 2/ 2 - sinx = sin I - sino - sin 31 - sinn 2 2 2 = 3=[ { :]-[9 8]-(125-3]-[2-53]) = 2 3 2 3 1-1- 2.25 +3-375 +5 -5-13 317/2 17 ) ) cosxdx * COSX 0.5 X 0.5+ 2 311 2 J-COSX A = 7/ 2 LOSX+ 31/ 2 o Cosxax A = Sinx - SinX 197/ 2 = Sin I - sino - sin 31 - sinn 2 2 2 = 347 ) y = x 3 ; [ 0 , 2 ] |4 9 ) y = x 2 + x + 1 j [ 2 , 3 ] A - J x 3 dx A= [ X 2 + X + 1 A = ( x 3 + x 2 + x ) 12 = [ "]-[4 ] = [2+ 2+ 3]- [3# + 2+27 1= 4 Discriminant : 55 ) ). x 2- 1 dx 2 b 2 - yac X - 1 = S2 ( x + 1) ( x-1) dx 1 - 3 NOX- int 10 1 x- inl 3 V + 2 x - in = S, ( x ti)dx = x + X 2 2+3 -[27 + 2 ] = 3 - 35
