Function [I] = Int2D(fxy,a,b,c,d,h) %UNTITLED Summary of...

function [I] = Int2D(fxy,a,b,c,d,h) %UNTITLED Summary of this function goes here % Detailed explanation goes here f=inline(fxy); isx=(b-a)/(2*h); isy=(d-c)/(2*h); A=zeros(isx,isy); for m=1:isy for n=1:isx A(n,m)=h^2/9*(f(a,c)+f(a+2*h,c)+f(a,c+2*h)+f(a+2*h,c+2*h)+4*(f(a+h,c)+f(a,c+h)+f(a+2*h,c+h)+f(a+h,c+2*h))+16*f(a+h,c+h)); a=a+2*h; end a=a-2*h*isx; c=c+2*h; end l=0; for m=1:isy for n=1:isx l=l+A(n,m) end end l end

Transcript

00:01 So, here according to the question, the integral given to us is integration from limits 9 to 16, 10 square root of x upon x minus 4 dx and we have to find its value using substitution method.

00:17 So, what we will do here, we will first of all let x is equal to t square.

00:22 Ok.

00:22 So, if x is equal to t square, then the value of dx will be equal to 2t multiplied by dt.

00:29 Ok.

00:29 So, we will change the limits too.

00:31 So, if x is equal to 9, then the value of t when x is equal to 9, the value of t will be 3.

00:40 Ok.

00:40 And when x is equal to 16, the value of t will be square root of 16 which is 4.

00:47 Ok.

00:48 So, the integral i will be now from limits 3 to 4 and 10 square root of x will be equal upon x minus 4 that means t square minus 4 and the value of dx is 2t dt.

01:02 Ok.

01:03 So, here it will become t square and 10 multiplied by 2 is 20.

01:07 So, we can take 20 outside because it is constant and integration 3 to 4, we have here t square dt upon t square minus 4.

01:16 So, now how will we solve it? 20 integration 3 to 4.

01:20 Now, we will write here t square, then we will write here minus 4 and plus 4.

01:26 So, minus 4 plus 4 will be again 0 and t square is as it is.

01:31 And here it is dt upon t square minus 4.

01:34 Now, we will separate the numerator and denominator like t square minus 4 upon t square minus 4 and plus 4 upon t square minus 4.

01:43 Ok.

01:43 So, the integration i value will be first of all 20, 20 integration from 3 to 4 and t square minus 4 upon t square minus 4 will be 1.

01:56 So, only integration of dt here.

01:58 Ok.

01:58 Then we have plus 20 and this 4 can be taken outside because it is constant.

02:05 So, 20 multiplied by 4 is 80 and integration from 3 to 4, we have dt upon t square minus we can write here 4 as 2 whole square.

02:16 Ok.

02:17 So, now the integration of dt is t.

02:19 So, 20 here t and limits are from 3 to 4 plus 80 here and the integration we know that the integration of dx upon x square minus a square.

02:31 Ok...

Question

Please give Ace some feedback