Am I doing this right? (Trapezoidal Rule)

Question

jennychan12 · Answer

$$\int\limits_{0}^{2} x^2dx$$
$$\frac{ 2 }{ 2(4) } (2)(1+4)$$
$$\frac{ 1 }{ 2 } (5) $$
$$= 2.5$$

abb0t · Answer

What is your "n" given value?

jennychan12 · Answer

oh sorry 
n=4

abb0t · Answer

So your midpoints (∆x) are:

[0,0.5][0.5,1][1,1.5][1.5,2]

b/c your given bounds are from 0 to 2.

jennychan12 · Answer

uhh, what midpoints?

abb0t · Answer

Sorry, subintervals.

jennychan12 · Answer

sorry, why would you need subintervals? cuz my textbook says nothing about that.

jennychan12 · Answer

ohhhh i see

abb0t · Answer

Then you just plug in those values into:

$$\int\limits_{a}^{b}f(x)dx \approx \frac{ ∆x }{ 2 }[f(x_0)+2f(x_1)+2f(x_2)+...=2f(x_{n-1})+f(x_n)]$$

jennychan12 · Answer

yeah i understand what i did wrong now....

abb0t · Answer

I hope this clears things up.

abb0t · Answer

Note that all the function evaluations, with the exception of the first and last, are multiplied by 2. Just so you can remember.

jennychan12 · Answer

i got the same answer when i divided it...

abb0t · Answer

ok. Whatever works. Best of luck :)

jennychan12 · Answer

oh just kidding my mistake..