AC6.2 - 6.2 CONCEPT QUESTIONS, page 417 1. To find I = f (...

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6 Integration 524 6.2 CONCEPT QUESTIONS, page 417 1. To find (() ) () I fgxgxd x = by the Method of Substitution, let () ug x = , so that du g x dx = . Making the substitution, we obtain I fud u = , which can be integrated with respect to u . Finally, replace u by ( ) x = to obtain the integral. 2. For 2 , x I xe dx = , we let 2 ux = − so that 2 du x dx = − or 1 2 . x dx du = − Then 1 2 u I ed u =− , which is easily integrated. But the substitution does not work for 2 x Je d x = because it does not reduce J to the form ( ) u , where f is easily integrable. EXERCISES 6.2, page 417 1. Put u = 4 x + 3 so that du = 4 dx , or dx = 1 4 du . Then 44 3 4 3 1 5 5 1 5 5 . xd xu d u u C x C += = + = + + z z 2. Let u = 2 x 2 + 1 so that du = 4 xdx. Then 42 1 2 1 27 7 1 8 8 1 8 28 xx d x ud u u C x C . = + = + + z z 3. Let u = x 3 3 x so that du = (3 x 2 3) dx . Then 32 2 2 3 33 11 (3 ) ( ) ) . x d x u d uu C x xC −− = = + = + ∫∫ 4. Put u = 2 x 3 x 2 + x so that du = (6 x 2 2 x + 1) dx . Then, 23 2 4 4 5 3 2 5 55 (6 2 1)(2 ) (2 ) x x xxx d x u d u uC C −+ = = + 5. Let u = 2 x 2 + 3 so that du = 4 xdx . Then 1 2 3 22 41 1 . (2 5) 2(2 5) x dx du u du u C C x === + = + ++ 6. Let u = x 3 + 4 x so that du = (3 x 2 + 4) dx . Then
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6 Integration 525 2 21 32 2 3 34 1 . (4 ) 4 xd u dx u du u C C xx u −− + == = + = + ++ 7. Put u = t 3 + 4 so that du = 3 t 2 dt or t 2 dt = 1 3 du . Then 2 3 1/2 3/2 3 22 33 ( 4 ) tt d t ud u u C t C += = + =+ + ∫∫ 8. Let 3 3 ut so that du = 3 t 2 dt . Then 23 3 / 2 3 / 2 5 / 2 3 5 / 2 55 3( 3 ) ( 3 ) . d t u d u u C t C = + = + + 9. Let u = x 2 1 so that du = 2 x dx and xdx du = 1 2 . Then, 29 9 1 0 0 11 1 0 2 0 (2 ) ) . x u du u C x C −= = + = + 10. Let u = 2 x 3 + 4 so that du = 6 x 2 dx or x 2 dx = 1 6 du 4 4 5 3 5 1 63 0 3 0 4) 4) . x x u d u u C x C = + = + + 11. Let u = 1 x 5 so that du = 5 x 4 dx or x 4 dx = 1 5 du 4 5 5 2 2 ln ln 1 . 15 5 5 u dx u C x C xu =− + =− + 12. Let u = x 3 1 so that du = 3 x 2 dx or x 2 dx = 1 3 du x x dx du u ud u u C x C 2 3 12 3 1 1 3 1 3 2 3 2 3 1 X Z Y = + = + X Z Y z // . 13. Let u = x 2 so that du = dx . Then 44 4 4 4ln ln ln ( 2) 2 du dx u C u C x C = = − + . 14. Let u = x 3 3, so that du = 3 x 2 dx , and 1 3 du = x 2 dx .
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6 Integration 526 2 3 3 22 2 2l n l n 3 . 33 3 3 xd u dx u C x C xu == + = + 15. Let u = 0.3 x 2 0.4 x +2 .±Then ± du = (0.6 x 0.4) dx = 2(0.3 x 0.2) dx . 03 02 04 2 1 2 1 2 1 2 2 2 2 .. ln ln( . . ) . x xx dx u du u C x x C −+ X Z Y += ++ X Z Y 16. Let u = 0.2 x 3 + 0.3 x . Then du = (0.6 x 2 + 0.3) dx = 0.3(2 x 2 + 1) dx . 21 1 1 10 3 2 3 3 x dx u du u C x x C + + + = + + X Z Y X Z Y . . ln ln . . . 17. Let u = 3 x 2 1 so that du = 6 xdx , or = 1 6 du . Then 2 2 331 1 ln ln 3 1 . 31 6 2 2 u dx u C x C ==+ = + 18. I x dx = X Z Y 2 3 1 . Let u = x 3 3 x + 1. Then, du = (3 x 2 3) dx = 3( x 2 1) dx , or ( x 2 1) dx = 1 3 du . Therefore, Iu d uu Cx x C + = + X Z Y + 1 3 1 3 1 3 13 ln ln . 19. Let u = 2 x so that du = 2 dx or dx = 1 2 du . Then 3 2 3.
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This note was uploaded on 03/26/2011 for the course ACCOUNTING 102 taught by Professor Albert during the Spring '11 term at Karachi Institute of Economics & Technology.

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AC6.2 - 6.2 CONCEPT QUESTIONS, page 417 1. To find I = f (...

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