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A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 19, 58, 78, 94, 105, 117, 129, 62, 40, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver. V = cm3

Respuesta :

Answer:

[tex]V \approx 1113[/tex]

Step-by-step explanation:

Here we have

Space [tex]x[/tex] in cm          Area in square centimeters  - [tex]A(x)[/tex]

  1. [tex]x = 0[/tex]                     [tex]0[/tex]
  2. [tex]x = 1.5[/tex]                 19
  3. [tex]x = 3[/tex]                   58
  4. [tex]x = 4.5[/tex]                78
  5. [tex]x = 6[/tex]                   94
  6. [tex]x = 7.5[/tex]               105
  7. [tex]x = 9[/tex]                  117
  8. [tex]x = 10.5[/tex]              129
  9. [tex]x = 12[/tex]                  62
  10. [tex]x = 13.5[/tex]                40
  11. [tex]x=15[/tex]                     0

As we can see, [tex]0 \leq x \leq 15[/tex] and we need to divide it into 5 subintervals, where

                                         [tex]\Delta x = \frac{15-0}{5} = 3[/tex]

is the length of each interval. Therefore, the subintervals are

                                 [tex][0,3], [3,6], [6,9], [9,12], [12,15][/tex]

The midpoint of the first interval is calculated as

                                        [tex]x^*_1 = \frac{3+0 }{2} = 1.5[/tex]

Similarly, we obtain that the midpoints of other subintervals are

                              [tex]x^*_2 =4.5, x^*_3= 7.5, x^*_4= 10.5, x^*_5 = 13.5[/tex]

By the Midpoint rule, we have

                      [tex]V \approx \sum_{i=1}^{5} A(x^*_i) \Delta x \\\\\phantom{V} = \Delta x (A(1.5) + A(4.5) + A(7.5) + A(10.5)+ A(13.5))[/tex]

From the given data above, we have

         [tex]A(1.5) = 19, A(4.5) = 78, A(7.5) = 105, A(10.5) = 129, A(13.5) = 40[/tex]

Therefore,

                      [tex]V \approx 3 (19+ 78+ 105+129+40 ) =[/tex] [tex]1113[/tex]

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