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anonymous

  • 5 years ago

represent ln5 as a definite intergral of the reciporcal function y=1/t

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{5} 1/t \]

  2. anonymous
    • 5 years ago
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    how would i grapth that to show the area it represents

  3. anonymous
    • 5 years ago
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    well you just need to graph 1/t and the area where x goes from 1 to 5

  4. anonymous
    • 5 years ago
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    how do i enter that in the calc

  5. anonymous
    • 5 years ago
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    with a graphic one you can do it. but I never used one

  6. anonymous
    • 5 years ago
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    do you know how 1/t looks like?

  7. anonymous
    • 5 years ago
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    no :/

  8. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=graph+1%2Fx

  9. anonymous
    • 5 years ago
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    the area you need is the one under the graph from x=1 to x=5

  10. anonymous
    • 5 years ago
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    so the first one?

  11. anonymous
    • 5 years ago
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    practice makes perfect, if you do calculus you need to be able to draw basic graphs of functions

  12. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=graph+1%2Fx+from+1+to+5

  13. anonymous
    • 5 years ago
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    so that is the the area? lol do i need to shade anything

  14. anonymous
    • 5 years ago
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    shade under the graph until the x axes

  15. anonymous
    • 5 years ago
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    perfect thanks

  16. anonymous
    • 5 years ago
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    what is the x-intercept of the tangent to y=e^x at x=45

  17. anonymous
    • 5 years ago
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    hmm, the derivative of e^x is e^x so the tangent at x=45 has slope of e^45 and the actual value is the same so then tangent line is y=e^45+e^45x the x intercept means that y=0 so 0=e^45+e^45x x=-1 that means that from x=45 you need to go back 1 to get the y intercept solution: x=44

  18. anonymous
    • 5 years ago
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    okay thats what i got. sweet

  19. anonymous
    • 5 years ago
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    well done!!

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