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AC6.2 - 6.2 CONCEPT QUESTIONS page 417 1 To find I = f g x...

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6 Integration 524 6.2 CONCEPT QUESTIONS, page 417 1. To find ( ( )) ( ) I f g x g x dx = by the Method of Substitution, let ( ) u g x = , so that ( ) du g x dx = . Making the substitution, we obtain ( ) I f u du = , which can be integrated with respect to u . Finally, replace u by ( ) u g x = to obtain the integral. 2. For 2 , x I xe dx = , we let 2 u x = − so that 2 du x dx = − or 1 2 . xdx du = − Then 1 2 u I e du = − , which is easily integrated. But the substitution does not work for 2 x J e dx = because it does not reduce J to the form ( ) f u du , 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 4 4 3 4 3 4 4 1 5 5 1 5 5 ( ) ( ) . x dx u du u C x C + = = + = + + z z 2. Let u = 2 x 2 + 1 so that du = 4 x dx. Then 4 2 1 2 1 2 7 7 1 8 8 1 8 2 8 x x dx u du u C x C ( ) ( ) . + = = + = + + z z 3. Let u = x 3 3 x so that du = (3 x 2 3) dx . Then 3 2 2 2 3 3 3 1 1 3 3 ( 3 ) (3 3) ( 3 ) . x x x dx u du u C x x C = = + = + 4. Put u = 2 x 3 x 2 + x so that du = (6 x 2 2 x + 1) dx . Then, 2 3 2 4 4 5 3 2 5 1 1 5 5 (6 2 1)(2 ) (2 ) x x x x x dx u du u C x x x C + + = = + = + + 5. Let u = 2 x 2 + 3 so that du = 4 x dx . Then 3 2 1 2 2 3 3 2 2 4 1 1 . (2 5) 2(2 5) x dx du u du u C C x u 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 2 1 3 2 2 3 3 4 1 . ( 4 ) 4 x du dx u du u C C x x u x x + = = = − + = − + + + 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 3/ 2 2 2 3 3 3 4 ( 4) t t dt u du u C t C + = = + = + + 8. Let 3 3 u t = + so that du = 3 t 2 dt . Then 2 3 3/2 3/ 2 5/ 2 3 5/2 2 2 5 5 3 ( 3) ( 3) . t t dt u du u C t C + = = + = + + 9. Let u = x 2 1 so that du = 2 x dx and xdx du = 1 2 . Then, 2 9 9 10 2 10 1 1 1 2 20 20 ( 2) ( 2) . x x dx 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 . Then 2 3 4 4 5 3 5 1 1 1 6 30 30 (2 4) (2 4) . x x dx u du u C x C + = = + = + + 11. Let u = 1 x 5 so that du = 5 x 4 dx or x 4 dx = 1 5 du . Then 4 5 5 2 2 2 2 ln ln 1 . 1 5 5 5 x du dx u C x C x u = − = − + = − + 12. Let u = x 3 1 so that du = 3 x 2 dx or x 2 dx = 1 3 du . Then x x dx du u u du u C x C 2 3 1 2 1 2 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 4 4 4 4 4ln ln ln ( 2) 2 du dx u C u C x C x u = = + = + = + . 14. Let u = x 3 3, so that du = 3 x 2 dx , and 1 3 du = x 2 dx .