Across-section of an airplane wing is shown. measurements of the thickness of the wing, in centimeters, at 16-centimeter intervals are 6.1, 19.9, 26.7, 29.0, 27.2, 27.5, 23.6, 20.9, 15.8, 9.1, and 3.2. use the midpoint rule with n = 5 to estimate the area of the wing's cross-section if a = 160. (assume the thickness of the edges is nonzero.)
answer has to be in cm^3

cross sectional area of the wing's is = 3404.8 cm²

Step-by-step explanation:

using n= 5 to estimate area of the wing's

a = 160

taking sum of thickness at n = 1, 3, 5, 7, 9

so sum of the measurement of the thickness at the given position

19.9 +29.0 + 27.5 +20.9 + 9.1 = 106.4

so the thickness is 106.4/5

           = 21.28 cm

cross sectional area of the wing's is = 160 × 21.28

                                                            = 3404.8 cm²

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

hope this !

let me know