Calculus: Early Transcendentals, by Anton, 7th Edition,ch06

Calculus - Early Transcendentals

This preview shows pages 1–4. Sign up to view the full content.

223 CHAPTER 6 Integration EXERCISE SET 6.1 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 + 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 ; 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 2 n + cos(( n 1) 2 n )+cos( 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 n , n n 2 n 1 n , 2; using right endpoints, A n = · n n + n n + + 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 ³ π 2 + π n ´ +cos µ π 2 + 2 π n + µ π 2 + ( n 1) π n ³ π 2 ´ ¸ π n n 2 5 10 50 100 A n 1 . 99985 1 . 93376 1 . 98352 1 . 99936 1 . 99985

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
224 Chapter 6 7. Endpoints 0 , 1 n , 2 n ,..., n 1 n , 1; using right endpoints, A n = s 1 µ 1 n 2 + s 1 µ 2 n 2 + ··· + s 1 µ 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 , 4 n 2( n 1) n , 1; using right endpoints, A n = s 1 µ n 2 n 2 + s 1 µ n 4 n 2 + + s 1 µ n 2 n 2 2 n n 2 5 10 50 100 A n 1 1 . 423837 1 . 518524 1 . 566097 1 . 569136 9. 3( x 1) 10. 5( x 2) 11. x ( x +2) 12. 3 2 ( x 1) 2 13. ( x + 3)( x 1) 14. 3 2 x ( x 2) 15. The area in Exercise 13 is always 3 less than the area in Exercise 11. The regions are identical except that the area in Exercise 11 has the extra trapezoid with vertices at (0 , 0) , (1 , 0) , (0 , 2) , (1 , 4) (with area 3). 16. (a) The region in question is a trapezoid, and the area of a trapezoid is 1 2 ( h 1 + h 2 ) w . (b) From Part (a), A 0 ( x )= 1 2 [ f ( a )+ f ( x )]+( x a ) 1 2 f 0 ( x ) = 1 2 [ f ( a f ( x x a ) 1 2 f ( x ) f ( a ) x a = f ( x ) 17. 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. 18. If the plane is rotated about the line y = x then A becomes B and vice versa. EXERCISE SET 6.2 1. (a) Z x x 2 dx = p x 2 + C (b) Z ( x +1) e x dx = xe x + C 2. (a) d dx (sin x x cos x + C )=cos x cos x + x sin x = x sin x (b) d dx µ x 1 x 2 + C = 1 x 2 + x 2 / 1 x 2 1 x 2 = 1 (1 x 2 ) 3 / 2 3. d dx h p x 3 +5 i = 3 x 2 2 x 3 so Z 3 x 2 2 x 3 dx = p x 3 +5+ C
Exercise Set 6.2 225 4. d dx · x x 2 +3 ¸ = 3 x 2 ( x 2 +3) 2 so Z 3 x 2 ( x 2 2 dx = x x 2 + C 5. d dx £ sin ( 2 x = cos (2 x ) x so Z cos (2 x ) x dx = sin ( 2 x ) + C 6. d dx [sin x x cos x ]= x sin x so Z x sin xdx = sin x x cos x + C 7. (a) x 9 / 9+ C (b) 7 12 x 12 / 7 + C (c) 2 9 x 9 / 2 + C 8. (a) 3 5 x 5 / 3 + C (b) 1 5 x 5 + C = 1 5 x 5 + C (c) 8 x 1 / 8 + C 9. (a) 1 2 Z x 3 dx = 1 4 x 2 + C (b) u 4 / 4 u 2 +7 u + C 10.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 01/24/2008.

Page1 / 55

Calculus: Early Transcendentals, by Anton, 7th Edition,ch06...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online