Question

using the trapezoidal rule. Just need to see where i went wrong..
\[\int\limits_{0}^{2} x^3 dx\]
over 4 intervals so. n = 4
\[h = \frac{ b-a }{ n } = \frac{ 2-0 }{ 4 }= \frac{ 1 }{ 2 }\]
\[T = \frac{ h }{ 2 } (y_0 + 2y_1 + 2y_2 +..... 2y_{n-1} + y_n)\]
\[T = \frac{ \frac{ 1 }{ 2 } }{ 2 } (0 + 2(1) + 8) = \frac{ 1 }{ 4 } (10) = \frac{ 10 }{ 4 }\]
Book answer is 17/4.
So I did some mistake... but I don't see what I did wrong.

Answer 1

Why do you only have \(3\) terms for \(y_0+y_1+y_2+y_3+y_4\)?

Answer 2

oh, that's what I did wrong.

Answer 3

\[T = \frac{ 1 }{ 4 }( 0 + 2(1) + 2(8) + 2(27) + 64)\]
?

Answer 4

I thought it was only three terms, because it's x = 0 to x = 2. so 0 , 1 , 2 .

Answer 5

why not do it by repeated applications A = h/2(f(a) + f(a + h))

A = 1/2(f(0) + f(1/2)) + 1/2(f(1/2) + f(1)) + 1/2(f(1) + f(3/2)) + 1/2(f(3/2)+ f(2))

Answer 6

I'm supposed to use the formula given from the book for this one.

Answer 7

\[
y_k = y(kh+a)
\]

Answer 8

well actually it is the formula is the book

height/2 times sum of 1st + last + 2 times middle

A = h/2[f(0) +2(f(1/2) + 2f(1) + 2f(3/2) + f(2)]

Answer 9

Oh, I think I see what I did wrong.
Since it's 4 sub intervals, so when x = 0, 0.5, 1,1.5, and 2.
\[T = \frac{ 1 }{ 4 } ( 0 + 2(\frac{ 1 }{ 8 }) + 2(1) + 2(\frac{ 27 }{ 8 }) + 8)\]
\[= \frac{ 1 }{ 4 }(\frac{ 2 }{ 8 } + 2 + \frac{ 54 }{ 8 } + 8)\]
\[= \frac{ 1 }{ 4 }\left( 17 \right)\]
\[ = 4.25\]