What M value should I use? (how do i find it)

Question

I was asked to calculate the area between two functions, y=e^x and y=x using midpoint riemann sums from 0-2 with n=4. so i found my answer by subtracting the second sum from the first.

The second question asks me to find a bound for the absolute error. I need to plug things into Em=M(b-a)^3/24n^2 but I don't understand if i should use e^x to find the max value of f''(x) or y=x? or both?

Answer 1

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

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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.