[Help~!!!][Calculus - Integrals]How can I 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 coudn't place integral symbol so I put 'S' instead. and 0 would be subscript (at below) and x would be superscript (at above) of integral 'S'

Thank you!!!

Drawing a sketch might help but I am supposed to prove this by algebraic expressions (although I think graph would be helpful if it helps me to understand better)

Thank you, ... Help me plz~!!!

Answer 1

Suppose that f(x) is an even function, then f(-x)=f(x)

Let g(x) = int^x_0 f(x) dx, then
g(-x) = int^(-x)_0 f(s) ds

Let s=-u so ds = -du then

g(-x) = - int^x_0 f(-u) du
= - int^x_0 f(u) du
= -g(x)

The function is odd.

Now suppose that f(x) is odd, f(-x)=-f(x) then
let g(x) = in^x_0 f(s) ds and then
g(-x) = int^-x_0 f(s) ds

Let s=-u then ds=-du so

g(-x) = -int^x_0 f(-u) du
= -int^x_0 -f(u) du
= int^x_0 f(u) du
= g(x)

so the function is even.

If you just look at a graph the result is counter-intuitive, but you have to remember that the bounds of integration matter!

Answer 2

Suggestion 1: Draw a sketch.

Suggestion 2: Using your notation, consider S(0 to -x) f(s)ds. Substitute s = -t, and so ds = -dt, and as s goes from 0 to -x, t goes from 0 to x, and so the integral is equal 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 means it's an odd function. (s and t are "dummy" variables -- it makes no difference what letter is used. The integral is the same if we write
S(0 to x) f(p) dp)

Likewise, if f is odd, f(-t) = -f(t) and so
S(0 to -x) f(s)ds = S(0 to x) f(t)dt, which shows it's an even function.

Suggestion 1 was more than a facetious holding exercise. Looking at the sketch you understand this apparently surprising result. Although with an even function both areas, equal in magnitude, are on the same side of the x axis and so we'd think they should be equal, integral from 0 to a negative value goes right to left instead of left to right, so the result of the integral has the opposite sign.

Answer 3

i'm no longer constructive if I understand the question yet i'll respond to it as I understand the question to be. If on the critical S from -x to x for f(s)ds, you get the respond to be 0, f(s) is strange. that is because on each and every area of the x-axis, f(s) will be doing an same ingredient and the y values will be purely the option. The y values will then upload jointly and cancel one yet another out. If on the critical S from x to -x you get an same fee as you do on 2*S from 0 to x for f(s)ds, then it really is a good function. that is because the y fee will be doing an same ingredient on an same area of the x axis on each and every area of the y axis. i'm hoping this permits. Calculus constructive is a discomfort.

Answer 4

This is easy. Function is even if f(-x) = f(x) and odd if f(-x) = - f(x). Define your integral as F(x) = S(from 0 to x)f(s)ds. Then F(-x) = S(from 0 to -x)f(s)ds = |make s = -t| = - S(from 0 to x)f(-t)dt = - S(from0 to x)f(t)dt = - F(x) therefore odd. Example: for f(x) = cos(x) even F(x) = sin(x) odd. Now for odd function f(-x) = -f(x). F(-x) = S(from 0 to -x)f(s)ds = |s = -t| = -S(from 0 to x)f(-t)dt = S(from 0 to x)f(t)dt = F(x) therefore even.