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Pre-Calc Homework Solutions 233

Pre-Calc Homework Solutions 233 - Section 6.1 9 1 x 3 233 6...

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9. [0.01, 5] by [ 2 3, 3] By setting the left endpoint at x 5 0.01 instead of x 5 0, we avoid an error that occurs when our calculator attempts to calculate NINT 1 } 1 x } , x , 1, 0 2 . The graph appears to be the same as the graph of y 5 ln x . 10. [ 2 5, 2 0.01] by [ 2 3, 3] By setting the right endpoint at x 5 2 0.01 instead of x 5 0, we avoid an error that occurs when our calculator attempts to calculate NINT 1 } 1 x } , x , 2 1, 0 2 . The graph appears to be the same as the graph of y 5 ln ( 2 x ). Section 6.1 Exercises 1. E ( x 2 2 2 x 1 1) dx 5 } x 3 3 } 2 x 2 1 x 1 C Check: } d d x } 1 } x 3 3 } 2 x 2 1 x 1 C 2 5 x 2 2 2 x 1 1 2. E ( 2 3 x 2 4 ) dx 5 x 2 3 1 C Check: } d d x } ( x 2 3 1 C ) 5 2 3 x 2 4 3. E ( x 2 2 4 ˇ x w ) dx 5 E ( x 2 2 4 x 1/2 ) dx 5 } x 3 3 } 2 } 8 3 } x 3/2 1 C Check: } d d x } 1 } x 3 3 } 2 } 8 3 } x 3/2 1 C 2 5 x 2 2 4 x 1/2 5 x 2 2 ˇ x w 4. E (8 1 csc x cot x ) dx 5 8 x 2 csc x 1 C Check: } d d x } (8 x 2 csc x 1 C ) 5 8 1 csc x cot x 5. E e 4 x dx 5 } 1 4 } e 4 x 1 C Check: } d d x } 1 } 1 4 } e 4 x 1 C 2 5 e 4 x 6. E } x 1 1 3 } dx 5 ln ) x 1 3 ) 1 C Check: } d d x } [ln ) x 1 3 ) 1 C ] 5 } x 1 1 3 } 7. E ( x 5 2 6 x 1 3) dx 5 } x 6 6 } 2 3 x 2 1 3 x 1 C 8. E ( 2 x 2 3 1 x 2 1) dx 5 } x 2 2 2 } 1 } x 2 2 } 2 x 1 C 9. E 1 e t /2 2 } t 5 2 } 2 dt 5 E ( e t /2 2 5 t 2 2 ) dt 5 2 e t /2 1 5 t 2 1 1 C 5 2 e t /2 1 } 5 t } 1 C 10. E } 4 3 } ˇ 3 t w dt 5 E } 4 3 } t 1/3 dt 5 t 4/3 1 C 11. E 1 x 3 2 } x 1 3 } 2 dx 5 E ( x 3 2 x 2 3 ) dx 5 } x 4 4 } 1 } x 2 2 2 } 1 C 5 } x 4 4 } 1 } 2 1 x 2 } 1 C 12. E 1 ˇ 3 x w 1 } ˇ 3 1 x w } 2 dx 5 E ( x 1/3 1 x 2 1/3 ) dx 5 } 3 4 } x 4/3 1 } 3 2 } x 2/3 1 C 13. E } 1 3 } x 2 2/3 dx 5 x 1/3 1 C 14. E (3 sin x 2 sin 3 x ) dx 5 2 3 cos x 2 } cos 3 3 x } 1 C 15. E } p 2
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