Question

@ganeshie8 please help!

Answer 1

@freckles can you please pleaese explain??

Answer 2

\[\int\limits_{a}^{b}f(x) \approx (b-a) \cdot \frac{f(a)+f(b)}{2}\]
I think that is the trapezoid approximation you are refering to

Answer 3

you are given a=1 and b=2
f(a)=f(1)=1
f(b)=f(2)=14

Answer 4

or you might be talking about this one:
\[\frac{b-a}{2 N}(fx_1)+2f(x_2)+2f(x_3)+ \cdots+2f(x_N)+f(x_{N+1}))\]

Answer 5

So your N+1=7 since that is how many function values you are given
So N=7-1=6

Answer 6

do you have any questions
I think you are looking for the second one there I mentioned

Answer 7

x_1=1
x_2=1.1
x_3=1.3
x_4=1.6
x_5=1.7
x_6=1.8
x_7=2

Answer 8

ahhh wait can you do the steps with me? i dont realy understand the trapezoidapproximation

Answer 9

it is just a pluggin in thing
doing using order of operations

Answer 10

\[\frac{b-a}{2 N}(f(x_1)+2f(x_2)+2f(x_3)+ \cdots+2f(x_N)+f(x_{N+1})) \\ \frac{2-1}{2(7-1)}(f(1)+2f(1.1)+2f(1.3)+2f(1.6)+2f(1.7)+2f(1.8)+f(2))\]

Answer 11

should i do the first or second one?

Answer 12

plug in and follow the order of operations

Answer 13

i assume the second one
but you tell me

Answer 14

which one is called the trapezoid approximation in your class

Answer 15

do they both come out with the same answer? ill look in my book

Answer 16

Not always I bet.

Answer 17

okay one second, ill check

Answer 18

actually that rule looks a little different from the second one

Answer 19

that one has no 2's in front of the x_i's between x_1 and x_(N+1)

Answer 20

i mean in front of the f(x_i)'s between f(x_1) and f(x_(N+1)

Answer 21

\[T_n=\frac{\Delta x }{2}(y_0+y_1+y_2+ \cdots+y_n)\]
So add up all the y values according to the rule you gave me
and then multiply that sum by (b-a)/n that is (2-1)/7

Answer 22

oops (2-1)/(7(2))

Answer 23

forgot about the 2 on bottom (you know below the delta x)

Answer 24

\[\frac{b-a}{2 N}(f(x_1)+2f(x_2)+2f(x_3)+ \cdots+2f(x_N)+f(x_{N+1})) \\ \frac{2-1}{2(7-1)}(f(1)+2f(1.1)+2f(1.3)+2f(1.6)+2f(1.7)+2f(1.8)+f(2)) \\ \frac{1}{2(6)}(1+2(3)+2(5)+2(8)+2(10)+2(11)+14) \\ \frac{1}{12}(1+6+7+16+20+22+14) \\ \frac{1}{12}(7+23+42+14) \\ \frac{1}{12}(30+56) \\ \frac{1}{12}(86) \\ \frac{86}{12}=\frac{43}{6} \approx 7.2\]
The other rule you provided should probably get something kinda close to this number

Answer 25

okay, thank you !!

Answer 26

so i guess we should have did 86/(7*2) here
86/14 if we want to use the same rule throughout your work