[Calculus - Integrals] Please~ help me to prove this!!!?

Question

Suppose f:[-a, a] --> R is either an even or an odd function. Show that S(from 0 to x) f(s)ds is therefore either an odd or an even function, respectively.

* I could put integral symbol here, so 'S' is integral symbol so 0 would be subscript and x would be superscript.

Thank you!!!

Answer 1

I'll call g(x) = (integral from 0 to x) f(s) ds.
We just need to see what g(-x) is.
g(-x) = (integral from 0 to -x) f(s) ds.
If f is even, then f(s) = f(-s), so this equals:
(integral from 0 to -x) f(-s) ds
and that is equivalent to
(integral from 0 to x) f(s) ds = g(x), so g is even.
Similarly, if f is odd, f(s) = -f(-s), so it equals
(integral from 0 to -x) -f(-s) ds
which is
(integral from 0 to x) -f(s) ds = -g(x), so g is odd.

Answer 2

I am not sure if I understand the question but I will answer it as I understand the question to be. If on the integral S from -x to x for f(s)ds, you get the answer to be 0, f(s) is odd. This is because on both sides of the x-axis, f(s) will be doing the same thing and the y values will be just the opposite. The y values will then add together and cancel each other out. If on the integral S from x to -x you get the same value as you do on 2*S from 0 to x for f(s)ds, then it is an even function. This is because the y value will be doing the same thing on the same side of the x axis on both sides of the y axis. I hope this helps. Calculus sure is a pain.

Answer 3

advice a million: Draw a caricature. advice 2: employing your notation, evaluate S(0 to -x) f(s)ds. replace s = -t, and so ds = -dt, and as s is going from 0 to -x, t is going from 0 to x, and so the critical is comparable to -S(0 to x) f(-t)dt If f is even, f(-t) = f(t), so S(0 to -x) f(s)ds = -S(0 to x) f(t)dt, which ability it rather is an unusual function. (s and t are "dummy" variables -- it makes no distinction what letter is used. The critical is the comparable if we write S(0 to x) f(p) dp) Likewise, if f is unusual, f(-t) = -f(t) and so S(0 to -x) f(s)ds = S(0 to x) f(t)dt, which shows it rather is a wonderful function. advice a million became extra effective than a facetious retaining exercising. observing the caricature you recognize this apparently superb consequence. inspite of the undeniable fact that with an excellent function the two areas, equivalent in fee, are on the comparable fringe of the x axis and so we would think of they could desire to be equivalent, critical from 0 to a unfavourable fee is going good to left as a replace of left to good, so the consequence of the critical has the alternative sign.