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2 < e < 3

  1. Nov 2, 2005 #1
    I have to carry out a few steps to show that 2 < e < 3. (e as in 2.71828..)

    Let f(t) = 1/t for t > 0.

    (a)
    Show that the area under y = f(t), above y = 0, and between t = 1 and t = 2 is less than 1 square unit. Deduce that e < 2.

    This is easy, just integrating and getting ln(2). From graph I see that ln(2) < 1 => e > 2.

    (b)
    Show that all tangent lines to the graph of f lie below the graph.

    Also not very hard; just showing that f'(t) < 0 gives that it's a decreasing function in the interval ]0,inf[, so the tangent lines have negative slopes, hence lies below the graph.

    (c)
    Find the lines T_2 and T_3 that are tangent to y = f(t) at t = 2 and t = 3, respectively.

    I got
    T_2: y = (-1/4)t + 1
    T_3: y = (-1/9)t + (2/3)

    (d)
    Find the area A_2 under T_2, above y = 0, and between t = 1 and t = 2. Also find the area A_3 under T_3, above y = 0, and between t = 2 and t = 3.

    I got
    A_2 = 5/8
    A_3 = 7/18

    (e)
    Show that A_2 + A_3 > 1 square unit. Deduce that e < 3.

    OK, so A_2 + A_3 = (5/8) + (7/18) = 73/72 > 1, but I can't see how this leads to e < 3. Any tips or suggestions?
     
  2. jcsd
  3. Nov 2, 2005 #2
    Oh, sorry. I guess I Didn't think long enough.

    int(t=1,t=3, 1/t) > A_2 + A_3 > 1, so ln(3) > 1, hence e < 3. bam!
     
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