
Andy C. answered 07/16/17
Math/Physics Tutor
The integral you are approximating is of exp(x) - x = e^x - x ,
which has a second derivative of exp(x) = e^x = e^x
Per excel, the function values are:
x y=f(x)=x y=g(x)=e^x g-f
0 0 1 1
0.25 0.25 1.284025417 1.034025417
0.5 0.5 1.648721271 1.148721271
0.75 0.75 2.117000017 1.367000017
1 1 2.718281828 1.718281828
1.25 1.25 3.490342957 2.240342957
1.5 1.5 4.48168907 2.98168907
1.75 1.75 5.754602676 4.004602676
2 2 7.389056099 5.389056099
0 0 1 1
0.25 0.25 1.284025417 1.034025417
0.5 0.5 1.648721271 1.148721271
0.75 0.75 2.117000017 1.367000017
1 1 2.718281828 1.718281828
1.25 1.25 3.490342957 2.240342957
1.5 1.5 4.48168907 2.98168907
1.75 1.75 5.754602676 4.004602676
2 2 7.389056099 5.389056099
---------------------------------------------------------------
M is the largest value of the second derivative in absolute value on the interval [0,2]
Per the table, M = 7.389056099, with a=0, b=2, and n = 4
So the error is 7.389056099 * (2-0)^3 / (24*4^2)
= 7.389056099 * 8/ (24*16)
= .15393866872916 = .15393866872916666....
The Riemann sum for y=x is 0.5*(0.25 + 0.75 + 1.25 + 1.75) = 2
Again, using excel, the Riemenn sum of exp(x) is 6.322985533;
Subtraction given an approximation of 4.322985533
The actual definite integral is exp(x) - x^2/2 on [0,2].
So the actual area is [exp(2)-2] - [ exp(0)-0] = exp(2) - 2 - 1
= exp(2)-3 = 4.3890561
Adding and Subtracting the error, in bold above, to and from
the approximation produces the interval:
(4.16904686427083333... , 4.4769242017291666...)
The actual area lies comfortably within this interval.