can someone show us the short way of doing the trapezodal rule (1 Viewer)

[red802]

i know how to do it the long way, e.g find 7 trapizia for something, and repeat the equation like 7 times, but how do u do the short way

 

[pLuvia]

Err, I don't know what you mean by the short way but here's the equation

b
∫ f(x) dx ≈ h/2 [f(a) + 2(x₁ + x₂ + x₃ +...+) + f(b)]
a

NOTE

h= b-a/ n

Where n is the number of sub intervals

 

[Riviet]

_a∫^b f(x) dx = h/2 . [y_o+y_n + 2(y₁+y₂+y₃+...+y_n-1)]

Where h is difference between consecutive x-values or (b-a)/n, y_o is the first y-value, y_n is the nth (last) y-value, and y₁ to y_n-1 represents all the other y-values in between the first and last y-value.

 

[whiteslice11]

a easy word way of remembering..

h/2 [(first plus last) + 2xrest]

yeh im lazy =)

 

[Riviet]

a easy word way of remembering..

h/2 [(first plus last) + 2xrest]

That's how I remember it.

The one I've given is the more formal way of writing it.

 

[pLuvia]

Opps forgot the brackets edited

 

[red802]

kool, can someone help me with this question because i done that an i got the wrong answer,

equation

f2,0(that intergal thing b=2, a =0)

(x^2+x)

this is how i done it

0.4/2 (0+2(0.56+1.44+2.64+4.16)+2)
=3.92

but the answer is 4.72
do u guys know the problem

 

[insert-username]

kool, can someone help me with this question because i done that an i got the wrong answer,

equation

f2,0(that intergal thing b=2, a =0)

(x^2+x)

this is how i done it

0.4/2 (0+2(0.56+1.44+2.64+4.16)+2)
=3.92

but the answer is 4.72
do u guys know the problem

Your problem is here:

0.4/2 (0+2(0.56+1.44+2.64+4.16)+2)

You haven't squared 2 and added 2 to it, like you have for each other value. 2² + 2 = 6. Use 6 instead of 2 and you'll get the right answer.


I_F

 

[red802]

thanks guys