# Thread: equation of curve that is a concatenation of sections of two curves

1. ## equation of curve that is a concatenation of sections of two curves

just need to verify this.
if a have a function f(x) and a function g(x) and the area under f(x)
from a known a to a variable x is (integral from a to x of)f(x);
let f1(x)= (integral from a to x of)f(x);
and
the area under g(x) from the same variable x to a known b is
(integral from x to b of)g(x)
let g1(x)=(integral from x to b of)g(x);
now let c(x)= f1(x)+g1(x);
first isnt c the area of the two regions
and isnt
c'(x)(the derivative of c) the function of the curve above the two
regions(i.e region under f1(x) and region under g1(x))
between a and b?
Or at least if the function of the curve above the two
regions(i.e region under f1(x) and region under g1(x))
is fn(x)
isnt fn(b)=c'(b);
by second fundamental theorem of calculus
would be greatful for any reply.thanks

2. Originally Posted by dynamo
just need to verify this.
if a have a function f(x) and a function g(x) and the area under f(x)
from a known a to a variable x is (integral from a to x of)f(x);
let f1(x)= (integral from a to x of)f(x);
and
the area under g(x) from the same variable x to a known b is
(integral from x to b of)g(x)
let g1(x)=(integral from x to b of)g(x);
now let c(x)= f1(x)+g1(x);
first isnt c the area of the two regions
Yes.

and isnt
c'(x)(the derivative of c) the function of the curve above the two
regions(i.e region under f1(x) and region under g1(x))
between a and b?
Yes.

Or at least if the function of the curve above the two
regions(i.e region under f1(x) and region under g1(x))
is fn(x)
isnt fn(b)=c'(b);
by second fundamental theorem of calculus
would be greatful for any reply.thanks
Yes, to all of those questions.

3. Originally Posted by HallsofIvy
Yes.

Yes.

Yes, to all of those questions.
thank you so much for replying to this question.This is very important to me but i have to ask another question.Are you absolutely sure beyond all possible doubt?!!in
$\displaystyle fn(b)=c'(b);$
b is from here
$\displaystyle let g1(x)=(integral from x to b of)g(x);$
just need to be absolutely sure.Thanks again.