CPhill

I don't know what they are asking for in (a)  or (b), GM.....but I know the others ....

(c)  With left endpoints, we have

                                     7

Width of  subinterval *  ∑   f ( x) 

                                    x =  4

 (d) The area with Left Endpoints is :

Width of  subinterval  * [ f(4) + f(5) + f(6) + f(7) ]  =

         (1)  * [ 17  + 26  + 37 + 50 ] =

                   130 units^2

(e)   With right endpoints, we have

Width of Interval  * [ f(5) + f(6) + f(7) + f(8)  ]  =

        (1)   *  [26 + 37 + 50 + 65 ]  =

                    178 units^2

 The actual area  is  [hide your eyes if you haven't had Calculus  ]

8                                                  8

 ∫  x^2 + 1   dx   =     [  x^3/3 + x ]       =   [ 8^3/3 + 8 ] - [ 4^4/3 + 4 ]  =  460 / 3   ≈  

4                                                 4

153.33 units^2

(f)  The Left Enpoints underestimate the area

(g)  The Right Enpoints overestimate the area

(h)  Trapezoidal method  =

[ b - a ]

______  *  [ f(4) + 2f(5) + 2f(6) + 2f(7) +  f(8)  ]

   2n

Where b = 8     a  = 4   and n = number of trapezoids = 4

So we have

[ 8 - 4]

______  * [ 17 + 52  + 74 + 100 + 65 ]     =    

   2 * 4

(1/2) * [   308 ] =

154  units^2

Note that this is very close to the actual area   !!!

(a)  

    12

    ∑    -7 - 3n

n  = 1

Note that other forms are possible.....

(b)  The sum can be found as

[ First Term + Last Term ] *  [ Number of terms] / 2

The number of terms is   [  43 - 10 ] / 3  + 1 =   33/3 + 1 =  11 + 1 = 12

So....the sum is

[ -10 + -43 ] * [ 12 / 2]  =

] -53] * 6  =

-318

x^2 + x  -  1  =  0

Rearrange as  :

x^2  =  1 - x      (1)

And

x^4 - 3x^2 + 3   =  

(x^2)^2 - 3x^2 + 3     (2)

Sub (1)  into (2)     and we have that

(1 - x)^2 - 3 (1 - x) +3

x*2 - 2x + 1 - 3 + 3x + 3

x^2 + x  - 2 + 3

( x^2 + x - 1 )  - 1 + 3

( x^2 + x - 1 ) + 2

0  +  2   =

2

The derivative  is  >   0   on these intervals : 

(-infinity, -3)  U  ( -2, 0)  U  (3, infinity)

These are the intervals of increase

The derivative  is  <  0  on these intervals

(-3, -2)  U (0, 3)

These are the intervals of decrease

x^2 - 13x  + 1  = 0

The sum of the roots  =13

The product of the roots  = 1

But....if the product of the  roots is  1.... we can call one root x  and the other   1/x

And the sum of the roots  =  x  + 1/x  =  13

So

(x)^2 - 13(x) + 1  = 0

(1/x)^2 - 13 (1/x) + 1  = 0        add these

(x)^2 + (1/x)^2 - 13 ( x + 1/x)  + 2  = 0

(x)^2  + (1/x)^2 -13 (13)  +  2  = 0

(x)^2 + (1/x)^2  - 169 + 2  =  0

(x)^2  + (1/x)^2  -167  = 0

(x)^2 + (1/x)^2  =   167      square both sides

(x)^4  + 2 (x)^2 * (1/x)^2  +  (1/x)^2  =  27889      simplify

x^4 + 2 (x^2/x^2) +  1/x^4  = 27889

(x)^4 +  2  +  (1/x)^4   +  2  =  27889       subtract 2 from both sides

x^2 +  1/x^4   = 27887

Suppose 1/x = 2/(y + z) = 3/(z + x) = (x^2 - y - z)/(x + y + z).  What is (z - y)/x?

Using the first two equalities  we have that

(y  + z)  =  2x       (1)

And using the second two equalities we have that

2(z + x)  = 3(y + z)  ⇒  2(z + x)  =  3(2x) ⇒   2z = 4x   ⇒  z = 2x   (2)

Sub (1) into the last expression in the equality :

(x^2 - (y + z) )                 x^2 - 2x        x ( x - 2)          x - 2

____________  =        ________   =  _______  =    ____

(x + y + z )                       x + 2x            3x                   3

So  using this and the first expression we have that

x - 2               1

____  =       ____      cross-multiply

   3                 x

x^2 - 2x = 3

x^2 - 2x - 3  =  0     factor

(x - 3) ( x + 1)  =   0

So either x = 3   or    x  = -1

Which means that   z = 6    or  z  = -2

Let x  = -1    and z =  -2

(y + z) = 2x

( y - 2)  = -2

Which means that y  = 0

Note that....when x = -1 , z =  -2    we have that

1/x = 2/(y + z) = 3/(z + x) = (x^2 - y - z)/(x + y + z)

1/ -1  = 2 / (0 - 2)   =  3 / (-2 - 1)   =  [ (-1)^2 - 0 + 2 ] / [ -1 - 0 - 2]  =  

1/-1   =  2/-2  = 3/ -3  =  3/-3

-1  = - 1  = - 1  = - 1

So

(z - y )  / x   =

(-2 - 0) / -1   =

2

BTW......The same result would be found when x = 3  and z  = 6

Let  the total watermelons we start with  =  N

1st  hour sold  =  [   N/2 + 1/2]

Remaining  is  what we started with - number sold in 1st hour =   N - [ N/2 + 1/2 ]  =  (1/2)N -1/2

2nd hour sold =   (1/2)[ (1/2)N - 1/2] + 1/2 =  [ (1/4)N  + 1/4 ]

Remaining is   what we started with at the end of the first hour - what we sold in the second hour  =

[ (1/2)N - 1/2 ] -  [ (1/4)N + 1/4]  =  [ (1/4)N - 3/4 ]

In the third hour  we sold  

(1/2)  [ (1/4)N - 3/4 ]+ 1/2    =   [ (1/8)N + 1/8]

What we  have remaining is the what we started with at the end of the second hour - what we sold in the third hour  =[(1/4)N - 3/4]  - [ (1/8)N + (1/8) ] =   (1/8)N - 7/8

And the number remaining  = 6  ...so...

(1/8)N - 7/8  = 6      multiplty through by  8

N - 7  =  48

N  = 55   is what we started with

To simplify things........call the third integer 2N....so we have that

(2N-4) + (2N -2)) + (2N) + (2N + 2) + (2N + 4) + (2N + 6)   = 282

2N (6)  + 6  =  282      subtract 6 from both sides

12N  =  276   divide both side by 12

N =  23

So....the integers are

2(23) - 4  = 42   and then   44, 46, 48 , 50 , 52

3 :  6  :  4

Call Natnan's share   N

Alesha gets N + 18 

And Jordan gets  ( 4/3)N

And we know that

(N + 18) / [(4/3)N]  =  6 / 4          cross-multiply

4 ( N + 18)  = 6 (4/3)N

4N + 72  =  8N

72  =  4N

18  = N

So   Jordan gets   (4/3)N  = (4/3)(18)  =  24  pounds

2.    Using the Law of Cosines

6^2 = 5^2  + 5^2  - 2 (5)(5) cos BAC

36 = 50 - 50 cos BAC

[36 - 50]

_______     = cos BAC

     -50

14

___  =  cos BAC

50

 7

__  =  cos BAC

25

So  sin BAC  =  24 / 25

And angle  BOC  = twice angle BAC

So

cos (2 * BAC)  = cos (BOC) =   1 - 2sin^2 (BAC)

cos (BOC)  = 1 - 2 (24/25)^2  =   1 - 2 ( 576/625)  =  [ 625 - 2(576) ] / 625  = -527/625

And using the Law of Cosines again

6^2  =  r^2 + r^2  - 2(r)(r) cos (BOC)

36 = 2r^2 - 2r^2 [ -527 / 625 ]

36 = 2r^2  [ 1 + 527/625 ]

18  = r^2 [ 1 + 527/625 ]

18 = r^2  [1152 /625 ]

r^2 =  [625 * 18 ] / 1152

r^2  = 11250 / 1152

r^2 =  625 / 64

r = 25/8  = the circumradius

So   the semiperimeter of triangle   BOC  is   [ 25/8+ 25/8 + 6 ]/2   =  49/8

So...using Heron's Formula....the area of BOC is

sqrt  [ (49/8) ( 49/8 - 25/8) (49/8 - 25/8) (49/8 - 6) ]  =

sqrt [ (49/8) ( 24/8) (24/8) ( 1/8) ] =

(7 * 24 * 1)              168              21      

__________  =        ___    =     ___  =   2.625 units^2

   64                          64               8