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# This is important to me and need someone to solve this.

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Solve every part of this problem please! Nov 23, 2019

#1
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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   !!!   Nov 23, 2019
edited by CPhill  Nov 23, 2019
edited by CPhill  Nov 23, 2019
#2
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Hey, CPhill thanks for the help which I understand but you made a mistake in your work and skipped something in the problem. In your work, you solved for (d) but labeled it as (c) and you solved for (e) but labeled it as (d). So you skipped (c) in the original problem.

GAMEMASTERX40  Nov 23, 2019
#3
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You are correct, GM....just a simple mistake...let me fix that....!!!   CPhill  Nov 23, 2019