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ppt kalin revision

1. KALKULUS INTEGRAL Kelompok 1

2. DISUSUN OLEH Agnes Margareta Nainggolan : 4233530003 Cindy Injilia S : 4233230007 Elida Silaban : 4232530004 Khairunnisa Aqilah : 4233530007 Sesy Ophelia Tampubolon : 4233230016

3. Penyelesaian  

4. 11. x Penyelesaian:  

5. 12. Penyelesaian :  

6. Jawaban 13.Terhadap sumbu x  

7. 14. terhadap sumbu y Jawaban :  

8. 15. Jawaban : y=4/x y=4/x x=1 x=4 y=4 y=1  

9. 16. y= ,x=2,y=0; mengelilingi sumbu y Jawaban : V= -(g dy V= dy V= . V= ( ( V= (16-8) V= 8  

10. 17. Y = 4- (x 0),x=0,y=0 mengelilingi garis x=2 Jawaban: y = 4- untuk x V= dx g (x) = 4- h (x) = 2 a = 0 dan b = 2   V= dx V= dx V= dx V= dx V= [4x- + V= [4(0)- + ] – [4(2)- + ] V= [8 - + ] V=  

11. 18. x= +1,y=2,x=0,y=0 mwngelilingi sumbu x Jawaban: x = +1, y=2, x=0, y=0 V = dy V = V = + V= π [ +4/3 √(2(0)) +0] - [ +4/3 √(2(2)) +2] V= π[4+8/3 √2 +2] V= 2 π  

12. 19. x = , y = 2, x = 0 ; mengelilingi garis y = 2 Jawaban : X = (y + 2 = y + 2 Y = – 2 V = – 2 dx = + 4 ) dx = - + 4x = [ . 16 - . 8 + 16 ) – ( 0 ) ] = + 16 ] = ] = 2 satuan volume = satuan volume  

13. 20. y = , y = x + 3 ; mengelilingi sumbu X Jawaban : = = = x + 3 – x – 3 = 0 ( x – 3 ) ( x + 1 ) = 0 X = 3 x = -1 y = 6 y = 2 v = - ) dx = ( x + 3 – ( ) dx = + 6x + 9 ) – ( = + + 6x + 9 ) ] dx = + + + 9x  

14. = + ( + 3 ( + 9 (3) ] – + (- + 3 ( + 9 ( -1 ) ] = + 9 + 27 + 27 ) – ( - + 3 – 9 ) ] = - – 6 ) ] = + 63 - + + 6 ) = ( 69 - + ) = ( 69 – 48 ) = 21  

15. 1. f(x)= + dari x=2;x=3 f’(x)= + f’(x)= + = + ( ) = - =   s dx dx dx dx dx dx   dx dx dx ) - = -   = ( . 81 -) . 16 -) = - - = - = =    

16. 2. Y=f(x) pada integral [a, b] Jawaban : s dx 24xy= + 48, dri x=2 sd x=4 y= y= + y= + y= + y’= + 2 . (-1) = - 2 =   S = = dx = dx dx dx dx + dx + dx + 2 dx  

17. = - 2 = - = ( . 64 -) . 8 -) = - - + 1 = + = =  

18. TERIMA KASIH