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Pre-Calc Homework Solutions 250

Pre-Calc Homework Solutions 250 - 250 Section 6.3 x ln x x...

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4. The graph of y 2 5 x ln x 2 x appears to be a vertical shift of the graph of y 1 5 E x 1 ln t dt (down 1 unit). Thus, y 2 appears to be an antiderivative of ln x which supports x ln x 2 x 1 C as the set of all antiderivatives of ln x . [0, 6] by [ 2 2, 5] Quick Review 6.3 1. } d d y x } 5 ( x 3 )(cos 2 x )(2) 1 (sin 2 x )(3 x 2 ) 5 2 x 3 cos 2 x 1 3 x 2 sin 2 x 2. } d d y x } 5 ( e 2 x ) 1 } 3 x 3 1 1 } 2 1 ln (3 x 1 1)(2 e 2 x ) 5 } 3 x 3 e 1 2 x 1 } 1 2 e 2 x ln (3 x 1 1) 3. } d d y x } 5 } 1 1 1 (2 x ) 2 } ? 2 5 } 1 1 2 4 x 2 } 4. } d d y x } 5 } ˇ 1 w 2 w 1 ( w x w 1 w 3 w ) 2 w } 5. y 5 tan 2 1 3 x tan y 5 3 x x 5 } 1 3 } tan y 6. y 5 cos 2 1 ( x 1 1) cos y 5 x 1 1 x 5 cos y 2 1 7. E 1 0 sin p x dx 5 2 } p 1 } cos p x 4 5 2 } p 1 } cos p 1 } p 1 } cos 0 5 2 } p 1 } ( 2 1) 1 } p 1 } 5 } p 2 } 8. } d d y x } 5 e 2 x dy 5 e 2 x dx Integrate both sides. E dy 5 E e 2 x dx y 5 } 1 2 } e 2 x 1 C 9. } d d y x } 5 x 1 sin x dy 5 ( x 1 sin x ) dx Integrate both sides. E dy 5 E ( x 1 sin x ) dx y 5 } 1 2 } x 2 2 cos x 1 C y (0) 5 2 1 1 C 5 2 C 5 3 y 5 } 1 2 } x 2 2 cos x 1 3 10. } d d x } 1 } 1 2 } e x (sin x 2 cos x ) 2 5 } 1 2 } e x (cos x 1 sin x ) 1 (sin x 2 cos x ) } 1 2 } e x 5 } 1 2 } e x cos x 1 } 1 2 } e x sin x 1 } 1 2 } e x sin x 2 } 1 2 } e x cos x 5 e x sin x Section 6.3 Exercises 1. Let u 5 x dv 5 sin x dx du 5 dx v 5 2 cos x E x sin x dx 5 2 x cos x 1 E cos x dx
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