Question

@DanJS

Answer 1

\[\sum_{i=1}^{6}f(a + i \Delta x )(\Delta x)\]

Answer 2

where
\[f(x) = x^3 +6x \] and
\[\Delta x = \frac{ b - a }{ n } = \frac{ 3 - 0 }{ 6 } = \frac{ 1 }{ 2 }\]

Answer 3

yup got that

Answer 4

n = 6

Answer 5

[a,b] = [0,3]

Answer 6

so you need to compute \[f(a+i \Delta x) = f(\frac{ i }{ 2 }) = (\frac{ i }{ 2 })^3 + 6*\frac{ i }{ 2 }\]

Answer 7

For i = 1,2,3,4,5,6

Answer 8

didnt we get -3.938?

Answer 9

then multiply that sum by (1/2)

Answer 10

no i think we had to recalculate it

Answer 11

let me see what this comes to ... one sec

Answer 12

okayy

Answer 13

omg it is x^3 - 6x

i was doing x^3 + 6x this whole time

Answer 14

oh!!

Answer 15

i didnt even notice!

Answer 16

well i got the same answer, -3.9

Answer 17

didnt you say that was wrong?

Answer 18

ok

Answer 19

i changed the bounds of the summation to i = 0 to i = N - 1so,

i = 0 to i = 5

Evaluating it i get, -8.43

Answer 20

and the exact answer from the integral of x^3-6x from 0 to 3 is -6.75

so for n=6 that is a decent approximation.