Question

Use the Trapezoidal Rule with n = 4 to approximate the integral from negative 1 to 0 of 1 over the square of the quantity x minus 1, dx.

Answer 1

https://gyazo.com/00654619772734b85d9d1d0e0bf61444

Answer 2

@jim_thompson5910 @Kenshin

Answer 3

Trapezoidal Rule:
\[\int_{a}^{b}f(x)dx = \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\]
where
\[\Large \Delta x = \frac{b-a}{n}\]

Answer 4

in this case
a = -1
b = 0
n = 4
\[\Large \Delta x = \frac{b-a}{n}\]
\[\Large \Delta x = \frac{0-(-1)}{4}\]
\[\Large \Delta x = \frac{1}{4}\]
\[\Large \Delta x = 0.25\]

Answer 5

If n = 4, then
\[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\]

Answer 6

I'm quite puzzled, especially with that last step

Answer 7

this?
\[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\]

Answer 8

Yes

Answer 9

do you recall from your lesson this formula?

\[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\]

btw I'm referencing this page

http://www.mathwords.com/t/trapezoid_rule.htm

Answer 10

sorry the formula is getting cut off, let me retry

Answer 11

Yea, I've been taking a look at that page too

Answer 12

\[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\]

Answer 13

that's the general formula for any integer n

if n were 4, then we have 5 terms we deal with, which are f(x0), f(x1), f(x2), f(x3), f(x4)

Answer 14

Ok, I think I see where you're going with this

Answer 15

we'll need the values of x0 through x4 they are...

\[\large x_0 = a = -1\]
\[\large x_1 = x_0 + \Delta x = -1+0.25 = -0.75\]
\[\large x_2 = x_1 + \Delta x = -0.75+0.25 = -0.5\]
\[\large x_3 = x_2 + \Delta x = -0.5+0.25 = -0.25\]
\[\large x_4 = x_3 + \Delta x = -0.25+0.25 = 0\]

-------------------------------------------------------

so in summary

\[\large x_0= -1\]
\[\large x_1= -0.75\]
\[\large x_2 = -0.5\]
\[\large x_3= -0.25\]
\[\large x_4 = 0\]

Answer 16

then you plug those values into \[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\] and you evaluate

Answer 17

I arrived at 1.0178, is that correct?

Answer 18

no

Answer 19

what are f(x0), f(x1), f(x2), f(x3), f(x4) equal to?

Answer 20

Just a moment

Answer 21

@BarfBag \[f(x)=\frac{ 1 }{ (x-1)^2 }\]
You would plug in each value given by @jim_thompson5910 to get what f(x0), f(x1), f(x2), f(x3), f(x4) are equal to.

Answer 22

f(x0) = 1/4, f(x1) = 0.327, f(x2) = 0.444, f(x3) = 0.64, f(x4) = 1

Answer 23

looks good @BarfBag

Answer 24

now compute `f(x0) + 2*f(x1) + 2*f(x2) + 2*f(x3) + f(x4)` and tell me what you get

Answer 25

I got 4.072

Answer 26

Multiply that by dx/2, correct?

Answer 27

correct

Answer 28

Multiply it by (1/4)/2

Answer 29

I ended up with B as my final answer

Answer 30

0.5089

Answer 31

B should be correct.

Answer 32

yep it's B) 0.5089

Answer 33

Thank you! That took over an hour to solve so I'm very appreciative of your help. XD