Sample 1E1 Solutions Sample 1E1 Solutions

    1. Two ways to do this. L'hopitals or by simplifying. If set x = 3 then both go to zero so can use l'hopital's which means we if we differentiate the top and the bottom the limit remains the same. So look at

      lim
      x® 3 
      		 1
      2x-5
      		 = 1
      1
      		 = 1.

    2. This is a telescoping series. Since
      1
      k(k+1)
      		 = 1
      k
      		 - 1
      k+1
      		 .
      This means that
      sk = 1- 1
      k+1
      so s¥ = 1.

    3. This is a harmonic form and so diverges, very slowly. Know this from p-series where
      1
      np
      converges for p > 1.

    1. If
      dy
      dx
      		 = x2-x
      then
      y = x3
      3
      		 - x2
      2
      		 +C.
      Since y(0) = 0 we know that C = 0 so
      y = x3
      3
      		 - x2
      2
      		 .
    2. Use integration by parts twice here.
      ó
      õ       		ex sinx dx = ex (-cosx) - ó
      õ       		ex (-cosx)dx = -ex cosx + ó
      õ       		ex cosx dx.
      Similarly
      ó
      õ       		ex cosx dx = ex sinx - ó
      õ       		ex sinx dx.
      Putting these together we get
      ó
      õ       		ex sinx dx = -ex cosx + ex sinx - ó
      õ       		ex sinx dx
      or (remembering to put in the constant now)
      ó
      õ       		ex sinx dx = 1
      2
      		 ex(sinx - cosx) + C.
    3. Here we use substitution. Taking u = 5x3+6 we have
      du
      15
      		 = x2dx.
      Using this we rewrite out integral as
      ó
      õ       		4
      15
      		 u-3du = [ 4
      15
      		 u-2
      -2
      		 ]x = 2x = 5,
      which in terms of x is
      -2
      15
      		 [(5(5)3+6)-2-(5(3)3+6)-2].
    4. We first rewrite this using partial fractions before integrating the various terms we get from this.

      So
      x-4
      (x-1)(x2+1)
      		 = A
      x-1
      		 + Bx+C
      x2+1
      and we want to determine A, B and C. To do this we mutiply across by (x-1)(x2+1) to get
      x-4 = A(x2+1)+(Bx+C)(x-1)
      which means
      -4
      =
      A-C,
      1
      =
      -B+C,
      0
      =
      A+B.
      From this we can see that
      A
      =
      -3
      2
      		 ,
      B
      =
      3
      2
      		 ,
      C
      =
      5
      2
      		 .
      Using this we have to integrate
      -3
      2
      		 1
      x-1
      		 +
      3
      2
      		 x+ 5
      2
      x2+1
      		 ,
      which can be simplified to
      - 3
      2
      		 1
      x-1
      		 + 3
      2
      		 x
      x2+1
      		 + 5
      2
      		 1
      x2+1
      		 .
      The first two terms are just ln terms which can be shown using substitution while the last one is tan-1 x. So our answer is
      - 3
      2
      		 ln | x-1 | + 3
      4
      		 ln(x2+1) + 5
      2
      		 tan-1 x + C.


    1. W = ó
      õ       		F(x)dx,
      where F = (5-0.05x)g. So
      W = ó
      õ       		1

      0 
      		 0 (5-0.05x)g dx = g[5x-0.05 x2
      2
      		 ]010 = 50g-0.05(50)g = 475J,
      taking g = 10ms-2.
    2. The line and curve intersect when we have x2+1 = 10-x or when x2+x-9 = 0. So we want
      x1 =
      -1±   ____
      Ö1+36
       
      2
      		 =
      -1+   __
      Ö37
       
      2
      since we are also bound by the y-axis. The area is then
      ó
      õ       		x1

      0 
      		 (10-x)-(x2+1) dx = [10x-x2/2-x3/3-x]0x1.
    3. This is a trigonometric substitution questions. We take x = 2tanq in which case dx = 2sec2qdq where we have 4+x2 = 4sec2q.

      Then we can rewrite our integral as
      ó
      õ       		2sec2qdq

      Ö
      22sec2q
      		 = ó
      õ       		secqdq = ln | secq+tanq | + C.
      We expressing this in terms of x we get
      ln |
        ____
      Ö4+x2
      2
      		 + x
      2
      		 | + C¢
      which equals
      ln |   ____
      Ö4+x2
       
      		 +x | + C¢¢,

    1. States that if have a,b and c where a £ b £ c and a and c both have the same limit then so does b. This lets us calculate limits for things we have difficulty with since we only need to show that a sandwich theorem applies to get them for example [sinx/x].
    2. We put the circle of interest in a square whose area we know. Then we randomly pick points in the square, noting whether they lie outside the cirle or not. The ratio of points in the circle to not is the same as the ratio of the circle area to the square area. In this way we can determine the area of the circle.

    1. Have f(x) = x3+x2+x+1, so
      df
      dx
      		 = 3x2+2x+1.
      Starting with x0 = 0 we have
      x1 = 0- 1
      1
      		 = -1
      and
      x2 = -1- 0
      2
      		 = -1.
      So we have already found a root at x = -1.
    2. Want to maximize the height. So find where slope is zero and 2nd derivative is negative.
      H = 5t-5t2
      so
      dH
      dt
      		 = 5-10t = 0,
      or t = 0.5.
      d2H
      dt2
      		 = -10 < 0
      so it is a maxima. At t = 0.5 we have H = 5/2-5/4 = 5/4 which is the max. height.

    1. The only issue is how many pieces we are going to use. I'll set n = 4. Recall for the Simpson's rule we have to take an even number since the basic unit involves 2 intervals. Once we have n = 4 we know that xi = 0.25i and yi = f(xi). We then just need to evaluate the yi's and plug it into the formula.
      y0
      =
      2
      0
      		 ,
      y1
      =
      7.87,
      y2
      =
      3.55,
      y3
      =
      1.87,
      y4
      =
      1.
      So flawed question since the first point is at ¥ in which case the integral diverges. Otherwise would have just plugged these values for the yi's into the simpson rule formula:
      h
      3
      		 (y0+4y1+2y2+4y3+y4).
    2. We don't have an error bound here since at x = 0 the first derivative is unbounded. If the integral was from 1 to 2 or something then this would not have been the case.

    1. If y = ex then lny = x and
      d
      dx
      		 lny = 1
      y
      		 dy
      dx
      		 = 1.
      Therefore
      dy
      dx
      		 = y
      and
      d
      dx
      		 ex = ex,
      a required.
    2. Again using the chain rule we have
      d
      dx
      		 lnxn = 1
      xn
      		 d
      dx
      		 xn = 1
      xn
      		 nxn-1 = n
      x
      		 .
      So
      d
      dx
      		 lnxn = n d
      dx
      		 lnx,
      which means that
      lnxn = nlnx + C.
      If x = 1 then we have
      ln1 = nln1 + C
      which implies that C = 0 since ln1 = nln1 = 0.
    3. This follows from the product rule. We have
      d
      dx
      		 uv = du
      dx
      		 v+u dv
      dx
      		 .
      Integrating both sides and using the fact that the integral is the antiderivative we get
      uv = ó
      õ       		du
      dx
      		 v+ ó
      õ       		u dv
      dx
      		 .
      Moving one of the integrals to the LHS we are done.
    4. We can express this as an integration by parts problem. The trick is to take [du/dx] = 1 in which case u = x and we have
      ó
      õ       		lnx = xlnx -x + C.

    1. We want to express a differentiable function f as a power series, or in other words to write it as
      f(x) = a0+a1(x-a)+a2(x-a)2+¼+an(x-a)n+¼.
      Then
      f¢(x) = a1+2a2(x-a)+3a3(x-a)2+¼+nan+1(x-a)n+¼.
      and
      f(k) = k!ak+¼+ n!
      k!
      		 an+k(x-a)n+¼.
      At x = a we have (x-a) = 0 so only the constant term in the above remain. This means that
      ak = f(k)
      k!
      		 ,
      which defines the coefficients of our power series which is the Taylor series.
    2. To represent x3/2 with a Taylor series we simply need to compute the coefficients which are given by
      ak = f(k)
      k!
      		 .
      So
      x3/2 = a3/2+ 3
      2
      		 a1/2(x-a)+¼+an(x-a)n+¼.
    3. To approximate 23/2 we pick a point to expand about, x = 1 say. Then using (x-a) = 1 in the above expansion we can get an approximate value for 23/2. Expanding about x = 2 we would need to know what root 2 was while around x = 1 we only have root 1 terms which are of course equal to one.

      Using only the first 3 terms we then have
      1+ 3
      2
      		 + 1
      2!
      		 3
      4
      		 = 1+ 3
      2
      		 + 3
      8
      		 = 23
      8
      		 .
      The exact value of 23/2 = 2.828.... while [23/8] = 2.875 which is not bad. If we wanted to be more accurate we could simply use more terms in our expansion.

    4. The interval of convergence would be the region in which the function is differentiable. The square root is only differentiable for x > 0, therefore if we expand about x = 1 our radius of convergence is 1. So the interval of convergence is 0 £ x £ 2.

File translated from TEX by TTH, version 2.70.
On 13 May 2002, 11:44.