Here is the situation.

area under 1/(x+1)

`Δx = (b − a)/n = (3 − 2)/4 = 0.25`

`y_0= f(a)`

` = f(2)`

`= 1/(2 + 1) = 0.3333333 `

`y_1= f(a + Δx) = f(2.25)` `= 1/(2.25+1) = 0.3076923`

`y_2= f(a + 2Δx) = f(2.5)` `= 1/(2.5+1) = 0.2857142`

`y_3= f(a + 3Δx) = f(2.75)` `= 1/(2.75+1) = 0.2666667`

`y_4= f(b) = f(3)` `= 1/(3+1) = 0.25`

So

Area ` = int_a^bf(x) text[d]x`

`approx 0.25/3 (0.333333+4(0.3076923)` `+2(0.2857142)+4(0.2666667)` `{:+0.25)`

`=0.2876831`

Notes

1. The actual answer to this problem is 0.287682 (to 6 decimal places) so our Simpson's Rule approximation has an error of only 0.00036%.

2. In this example, the curve is very nearly parabolic, so the 2 parabolas shown above practically merge with the curve `y=1/(x+1)`.

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