Approximate \int_0^2(1)/(1+x^3)dx using the Midpoint Rule, given n=4? Please check my work

f(\barx_1)=3/4
f(\barx_2)=11/36
f(\barx_3)=37/504
f(\barx_4)=93/3640

and \Deltax=(b-a)/n=(2-0)/4=1/2


\thereforeM_4=1/2(3/4+11/36+37/504+93/3640)

1 Answer
Apr 17, 2018

Please see below.

Explanation:

On interval [0,2] with n = 4, we get Deltax = (2-0)/4 = 1/2

so the subintervals are

[0,1/2], [1/2,1], [1,3/2], [3/2,2]

The corresponding midpoints are

1/4, 3/4, 5/4 and 7/4.

each of these has the form a/4 (a = 1, 3, 5, 7)

f((a/4)) = 1/(1+(a/4)^3) = 1/(1+(a^3/64)) = 1/((64+a^3)/64)

f(a/4) = 64/(a^3+64).

Your values in the comments are correct:

So we need:

(64/65+64/91+64/189+64/407) * 1/2 = (32/65+32/91+32/189+32/407)

65 = 5 * 13
91 = 7 * 13
189 = 9 * 21 = 7 * 27
407 = 11 * 37

The least common factor of the denominators is: 5 * 7 * 11 * 13 * 27 * 37

Now grind out the arithmetic. (And be happy you were born in the age of electronic calculators. Math students used to have to do this stuff all the time.)