Question

Proof that defined integral (top row = b and bottom row = a) f(x)dx = defined integral f(a+b-x)dx. Last question...

Answer 1

definite integral*

Answer 2

\[\int\limits_{a}^{b}f(x)dx = \int\limits_{b}^{a}f(x)(-dx)\]
\[\int\limits_{a}^{b}f(a+b-x)dx = \int\limits_{b}^{a}f(x)(-dx)=\int\limits_{a}^{b}f(x) dx\]
each time you switch the upper and lower limits, you need to flip the sign of dx

Answer 3

for example, when taking an integral from x=1 to x=3, each infinitesimal step dx will be positive
however, when taking an integral from x=3 to x=1, each infinitesimal step dx must be negative

Answer 4

actually, this problem can be done with u substitution: u=a+b-x, du=-dx

Answer 5

\[\int\limits_{a}^{b}f(a+b-x)dx=\int\limits_{u(a)}^{u(b)}f(u)(-du)=\int\limits_{b}^{a}f(u)(-du)=\int\limits_{a}^{b}f(x)dx\]

Answer 6

Thx! It's seems a lot more simple now!

Answer 7

glad this helped :)