AC6.6 - 6.6 CONCEPT QUESTIONS page 458 1 b a f x g x dx 2 b...

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6 Integration 561 6.6 CONCEPT QUESTIONS, page 458 1. [() () ] b a f xg x d x 2. [ ( ) ( )] [ ( ) ( )] [ ( ) ( )] bcd abc fx gxd x gx fxd x x −+−+ ∫∫∫ EXERCISES 6.6, page 458 1. −− = += z () ( ) ( ) xx d x 32 43 0 6 0 6 6 2 6 2 6 108 1 4 1 4 sq units. 2. = −= = z d xxx 45 0 2 0 2 28 1 2 1 5 32 5 8 5 sq units. 3. A x xd x x x x x d x =− + = z zz 11 2 1 22 0 1 1 0 212 0 1 / (by symmetry). Let u = 1 x 2 so that du = 2 x dx or x dx = - 1 2 du . Also, if x = 0, then u = 1 and if x = 1, u = 0. So 0 1 1/2 3/2 12 2 2 23 3 3 0 1 (2)( ) , or sq unit. A u du u = = 4. A x x dx x x dx x x dx x + + + = + =+ X Z Y z z 2 4 2 4 2 2 4 24 2 0 2 2 0 2 0 2 2 0 ln( ) = (ln 8 ln 4)2 = ln 4 sq units. 5. Ax x d x d x x x = − + + z z ( ) // 2 0 4 0 4 0 4 1 2 4 3 = 8 + 32 3 8 3 = sq units. 6. x d x x x d x x x x = + + z z [( ) ] ( )( ) 2 2 0 4 0 4 0 4 2 3 1 2 = 16 3 16 3 88 −+= .
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6 Integration 562 7. The required area is given by () // / / x x dx x x dx x x x x 21 3 1 0 13 2 3 43 1 0 3 0 1 0 1 1 3 3 4 3 4 1 3 −+ = + zz = −+−= ( ) 1 3 3 4 3 4 1 3 1 2 1 sq units. 8. Ax x d x xx d x =+ + + z z [( ) ( )] [( ) ] 66 1 2 3 0 2 4 0 + + z z [ ( ) ] 3 2 3 0 2 4 0 xd x x x d x = 3 4 2 4 0 1 2 2 1 4 4 0 2 6 6 12 24 2 12 4 22 x +++ = + + = ( ) ( ) ( ) sq units. 9. The required area is given by −− = =+= z x x 2 1 2 3 1 2 1 3 8 3 1 3 3 sq units. 10. d x x d x =− z z 22 0 2 2 2 42 4 =− + =−+ 24 2 8 1 3 8 3 3 0 2 ( ) 11. y = x 2 5 x + 4 = ( x 4)( x 1) = 0 if x = 1 or 4. These give the x- intercepts. x d x x x x + + z 2 1 3 3 5 2 2 1 3 1 3 54 4 =−+ − −− + − = = ( ) . 91 2 4 3 45 2 1 3 5 2 10 3 1 3 = 32 3 sq units .
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6 Integration 563 12. The required area is given by −= = + = z xd x x 34 1 0 1 0 1 4 1 4 1 4 1 4 01 () . 13. The required area is given by −−+ =+ =+ = z . / 1 9 18 27 2 3 32 0 9 0 9 x x x 14. Ax x d x x x =− + z // 1 2 1 4 2 3 12 2 0 4 0 4 15. −− = z ed x e xx (/ ) 2 4 2 -2 4 =−+ = 4 16 3 4 3 sq units. 2 21 ee sq units.
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6 Integration 564 16. Ax e d x x e d x xx =− − = −− z z 22 0 1 0 1 . Let u = x 2 so that du = 2 x dx or x dx = - 1 2 du . Also, if x = 0, then u = 0 and if x = 1 then u = 1. So Ae d u e uu =− z 1 2 1 2 0 1 0 1 17. d x =+ z [( ) ] 2 1 3 31 = () (). 96 2 1 3 38 3 +−+= 18. xd x z [( ) ( )] 24 2 1 2 + + 1 3 1 2 32 1 2 6 x 19. x x d x ++ + z 2 0 2 23 3 + + 1 3 3 2 1 3 3 2 0 2 4 (8) ( ) 20. The region is shown in the figure on the right. + = 1 2 1 2 1 2 11 1 ee . sq units = + z x x x 2 1 3 3 1 3 1 3 + + z d x 2 1 2 6 =− + + − + − = ( ) 8 3 1 3 1 2 1 2 2 12 6 16 sq units + z d x 2 0 2 3 =−= 6 8 3 10 3 sq units
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6 Integration 565 Ax x d x =− + z [( ) ( )] 92 3 2 1 1 = −− + 1 3 32 1 1 6 xx x 21. x d x =+ z [( ) ] 23
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AC6.6 - 6.6 CONCEPT QUESTIONS page 458 1 b a f x g x dx 2 b...

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