Question

Trapezoidal rule, in general. AM I CORRECT ?

Answer 1

\(\large\color{black}{ \displaystyle \int_{a}^{b}f(x)~dx\approx }\)


\(\large\color{black}{ \displaystyle \frac{\Delta x}{2}\cdot~{\huge [} f(x_0)+2f(x_1)+2f(x_3)+{\tiny~}...{\tiny~}+2f(x_{n-1})+f(x_n){\huge ]} }\)


\(\Huge(\)where \(\large\color{black}{ \displaystyle \Delta x =\frac{b-a}{n} }\), and \(\large{\rm n}\) is the number of trapezoids used \(\Huge)\)

Answer 2

Or,

\(\normalsize\color{black}{ \displaystyle \frac{(b-a)}{2n}\cdot~{\huge [} f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+2f(a+3\Delta x)+2f(a+4\Delta x)+ \\ {\tiny~}...{\tiny~}+2f(a+(n-1)\Delta x)+f(a+n\Delta x){\huge ]} }\)

Answer 3

i think both true

Answer 4

And that will be,

\(\normalsize\color{black}{ \displaystyle \frac{(b-a)}{2n}\cdot~{\huge [} f(a)+2f(a+\frac{b-a}{n})+2f(a+\frac{2(b-a)}{n})+2f(a+\frac{3(b-a)}{n})+ \\ ~ \\ \displaystyle 2f(a+\frac{4(b-a)}{n})+ \\ \displaystyle {\tiny~}...{\tiny~}+2f(a+\frac{(n-1)(b-a)}{n})+f(b){\huge ]} }\)

Answer 5

this is in terms of a b and n

Answer 6

I just try to play around if I find anything....

Answer 7

\(\small\color{black}{ \displaystyle \frac{(b-a)}{2n}\cdot~{\huge [} f(a)+2f\left(\frac{na+b-a}{n}\right)+2f\left(\frac{na+2b-2a}{n}\right)+2f\left(\frac{na+3b-3a}{n}\right)+ \\ ~ \\ \displaystyle 2f\left(\frac{na+4b-4a}{n}\right)+ \displaystyle {\tiny~}...{\tiny~}+2f\left(\frac{na+(n-1)(b-a)}{n}\right)+f(b){\huge ]} }\)

Answer 8

\(\small\color{black}{ \displaystyle \frac{(b-a)}{2n}\cdot~{\huge [} f(a)+2f\left(\frac{na+b-a}{n}\right)+2f\left(\frac{na+2b-2a}{n}\right)+2f\left(\frac{na+3b-3a}{n}\right)+ \\ ~ \\ \displaystyle 2f\left(\frac{na+4b-4a}{n}\right)+ \displaystyle {\tiny~}...{\tiny~}+2f\left(\frac{na+nb-na-b+a}{n}\right)+f(b){\huge ]} }\)

Answer 9

\(\small\color{black}{ \displaystyle \frac{(b-a)}{2n}\cdot~{\huge [} f(a)+2f\left(\frac{na+b-a}{n}\right)+2f\left(\frac{na+2b-2a}{n}\right)+2f\left(\frac{na+3b-3a}{n}\right)+ \\ ~ \\ \displaystyle 2f\left(\frac{na+4b-4a}{n}\right)+ \displaystyle {\tiny~}...{\tiny~}+2f\left(\frac{nb-b+a}{n}\right)+f(b){\huge ]} }\)

Answer 10

NO, it is more convenient to use the very first formula, just have to always keep in mind \(\Delta x = (b-a)/n\) (which makes sense logically)

Answer 11

Tnx for being here though....