What do I input to find the area of the region bounded by the hyperbola?

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Rachel Bowlin
Rachel Bowlin 2021 年 6 月 23 日
コメント済み: Prakhar Rai 2021 年 6 月 29 日
I am trying replicate the problem below in Matlab. What is the input I use?
Thanks!

回答 (1 件)

Prakhar Rai
Prakhar Rai 2021 年 6 月 23 日
By using definite integral over the intersection points of the two functions.
Area b/w two curves = Integral (f(x)-g(x)) ;
  2 件のコメント
Rachel Bowlin
Rachel Bowlin 2021 年 6 月 23 日
What am I doing wrong?
Prakhar Rai
Prakhar Rai 2021 年 6 月 29 日
Create the function f(x)= 3/2 sqrt(x^2-4),f(x)= - 3/2 sqrt(x^2-4)
The area b/w the functions would be Integral(f(x) from x [2:3]) + Integral(-f(x) from x [2:3]) as at x=2 hyperbola becomes 0 and x=3 the two functions cross.
Due to symmetry:
Integral(f(x) from x [2:3]) =Integral(-f(x) from x [2:3])
So just calculate Integral(f(x) from x [2:3]) and multiply by 2.
Twice since area is same both above and below x axis.
fun = @(x) 3./2 .* sqrt((x.^2)-4);
ans = 2 .* integral(fun,2,3);
You need to pass function handle in integral().
Also as mentioned in errors use '.^' and '.* ' instead of ^ and * .
Please go through this page: Integration to get complete understanding.
Please upovote if you found this helpful.

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