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chapter_5 - CHAPTER 5 Integration EXERCISE SET 5.1 1...

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225 CHAPTER 5 Integration EXERCISE SET 5.1 1. Endpoints 0 , 1 n , 2 n , . . . , n - 1 n , 1; using right endpoints, A n = " r 1 n + r 2 n + · · · + r n - 1 n + 1 # 1 n n 2 5 10 50 100 A n 0 . 853553 0 . 749739 0 . 710509 0 . 676095 0 . 671463 2. Endpoints 0 , 1 n , 2 n , . . . , n - 1 n , 1; using right endpoints, A n = n n + 1 + n n + 2 + n n + 3 + · · · + n 2 n - 1 + 1 2 1 n n 2 5 10 50 100 A n 0 . 583333 0 . 645635 0 . 668771 0 . 688172 0 . 690653 3. Endpoints 0 , π n , 2 π n , . . . , ( n - 1) π n , π ; using right endpoints, A n = [sin( π/n ) + sin(2 π/n ) + · · · + sin( π ( n - 1) /n ) + sin π ] π n n 2 5 10 50 100 A n 1 . 57080 1 . 93376 1 . 98352 1 . 99935 1 . 99984 4. Endpoints 0 , π 2 n , 2 π 2 n , . . . , ( n - 1) π 2 n , π 2 ; using right endpoints, A n = [cos( π/ 2 n ) + cos(2 π/ 2 n ) + · · · + cos(( n - 1) π/ 2 n ) + cos( π/ 2)] π 2 n n 2 5 10 50 100 A n 0 . 555359 0 . 834683 0 . 919405 0 . 984204 0 . 992120 5. Endpoints 1 , n + 1 n , n + 2 n , . . . , 2 n - 1 n , 2; using right endpoints, A n = n n + 1 + n n + 2 + · · · + n 2 n - 1 + 1 2 1 n n 2 5 10 50 100 A n 0 . 583333 0 . 645635 0 . 668771 0 . 688172 0 . 690653 6. Endpoints - π 2 , - π 2 + π n , - π 2 + 2 π n , . . . , - π 2 + ( n - 1) π n , π 2 ; using right endpoints, A n = cos - π 2 + π n + cos - π 2 + 2 π n + · · · + cos - π 2 + ( n - 1) π n + cos π 2 π n n 2 5 10 50 100 A n 1 . 57080 1 . 93376 1 . 98352 1 . 99936 1 . 99985
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226 Chapter 5 7. Endpoints 0 , 1 n , 2 n , . . . , n - 1 n , 1; using right endpoints, A n = s 1 - 1 n 2 + s 1 - 2 n 2 + · · · + s 1 - n - 1 n 2 + 0 1 n n 2 5 10 50 100 A n 0 . 433013 0 . 659262 0 . 726130 0 . 774567 0 . 780106 8. Endpoints - 1 , - 1 + 2 n , - 1 + 4 n , . . . , - 1 + 2( n - 1) n , 1; using right endpoints, A n = s 1 - n - 2 n 2 + s 1 - n - 4 n 2 + · · · + s 1 - n - 2 n 2 + 0 2 n n 2 5 10 50 100 A n 1 1 . 423837 1 . 518524 1 . 566097 1 . 569136 9. Endpoints - 1 , - 1 + 2 n , - 1 + 4 n , . . . , 1 - 2 n , 1; using right endpoints, A n = h e - 1+ 2 n + e - 1+ 4 n + e - 1+ 6 n + . . . + e 1 - 2 n + e 1 i 2 n n 2 5 10 50 100 A n 3 . 718281 2 . 851738 2 . 59327 2 . 39772 2 . 37398 10. Endpoints 1 , 1 + 1 n , 1 + 2 n , . . . , 2 - 1 n , 2; using right endpoints, A n = ln 1 + 1 n + ln 1 + 2 n + . . . + ln 2 - 1 n + ln 2 1 n n 2 5 10 50 100 A n 0 . 549 0 . 454 0 . 421 0 . 393 0 . 390 11. Endpoints 0 , 1 n , 2 n , . . . , n - 1 n , 1; using right endpoints, A n = sin - 1 1 n + sin - 1 2 n + . . . + sin - 1 n - 1 n + sin - 1 (1) 1 n n 2 5 10 50 100 A n 1 . 04729 0 . 75089 0 . 65781 0 . 58730 0 . 57894 12. Endpoints 0 , 1 n , 2 n , . . . , n - 1 n , 1; using right endpoints, A n = tan - 1 1 n + tan - 1 2 n + . . . + tan - 1 n - 1 n + tan - 1 (1) 1 n n 2 5 10 50 100 A n 0 . 62452 0 . 51569 0 . 47768 0 . 44666 0 . 44274 13. 3( x - 1) 14. 5( x - 2) 15. x ( x + 2) 16. 3 2 ( x - 1) 2 17. ( x + 3)( x - 1) 18. 3 2 x ( x - 2)
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Exercise Set 5.2 227 19. false; the area is 4 π 20. false; consider the left endpoint approximation on [1 , 2]. 21. true 22. true; a differentiable function is continuous 23. A (6) represents the area between x = 0 and x = 6; A (3) represents the area between x = 0 and x = 3; their difference A (6) - A (3) represents the area between x = 3 and x = 6, and A (6) - A (3) = 1 3 (6 3 - 3 3 ) = 63. 24. A (9) = 9 3 / 3 , A ( - 3) = ( - 3) 3 / 3, and the area between x = - 3 and x = 9 is given by A (9) - A ( - 3) = (9 3 - ( - 3) 3 ) / 3 = 252. 25. B is also the area between the graph of f ( x ) = x and the interval [0 , 1] on the y - axis, so A + B is the area of the square. 26. If the plane is rotated about the line y = x then A becomes B and vice versa. 27. The area which is under the curve lies to the right of x = 2 (or to the left of x = - 2). Hence f ( x ) = A 0 ( x ) = 2 x ; 0 = A ( a ) = a 2 - 4, so take a = 2 (or a = - 2 to measure the area to the left of x = - 2). 28. f ( x ) = A 0 ( x ) = 2 x - 1 , 0 = A ( a ) = a 2 - a , so take a = 0 (or a = 1). 30. Intuitively it is the area represented by a set of tall thin rectangles, stretching from x = a to x = b , each having height C ; in other words C ( b - a ). Analytically it is given by Z b a [( f ( x ) + C ) - f ( x )] dx = C ( b - a ).
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