Question

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`

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

Answer 1

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.)