I need to show that if f is integrable on [0,b] and f is odd then f is integrabl
ID: 1892551 • Letter: I
Question
I need to show that if f is integrable on [0,b] and f is odd then f is integrable on [-b,b]Explanation / Answer
If f is integrable on [-b,b] and f is an odd function, then look at it this way: Int_[-b,b] f(x) dx by additivity of the integral: = Int_[-b,0) f(x) dx + Int_[0,b] f(x) dx by using the change of variables x --> -x: = Int_[b,0) f(-x) (-1) dx + Int_[0,b] f(x) dx by using the fact that f is odd, and pulling the (-1) out front: = -Int_[b,0) -f(x) dx + Int_[0,b] f(x) dx by using the fact that Int_[a,b] f(x) dx = -Int_[b,a] f(x) dx: = Int_(0,b] -f(x) dx + Int_[0,b] f(x) dx by pulling the (-1) outside, and remembering that Int_(0,b] f(x) dx = Int_[0,b] f(x) dx; i.e. the value of the function at the point zero does not affect the integral: = -Int_[0,b] f(x) dx + Int_[0,b] f(x) dx since f is integrable, both of the terms are finite and the first is (-1) times the other: = 0 You can follow pretty much the same logic for the even. Try to work it out before reading this: Suppose now that f is integrable on [-b,b] and that f is even. Then: Int_[-b,b] f(x) dx by additivity of the integral: = Int_[-b,0) f(x) dx + Int_[0,b] f(x) dx by using the change of variables x --> -x: = Int_[b,0) f(-x) (-1) dx + Int_[0,b] f(x) dx by pulling the (-1) out front: = -Int_[b,0) f(-x) dx + Int_[0,b] f(x) dx by using the fact that f is an even function: = -Int_[b,0) f(x) dx + Int_[0,b] f(x) dx by using the fact that Int_[a,b] f(x) dx = -Int_[b,a] f(x) dx: = Int_(0,b] f(x) dx + Int_[0,b] f(x) dx by using that Int_(0,b] f(x) dx = Int_[0,b] f(x) dx; i.e. the value of the function at the point zero does not affect the integral: = Int_[0,b] f(x) dx + Int_[0,b] f(x) dx since f is integrable, both of the terms are finite and the same: = 2 Int_[0,b] f(x) dx (Edit: It is actually not in the definition of the integral that for any function f integrable on [a,b], Int_[a,c] f(x) dx + Int_[c,b] f(x) dx = Int_[a,b] f(x) dx This is a property that the integral should satisfy if it's going to be well-defined. This isn't hard to prove if you're using the Riemann integral; it's slightly harder if you're using the Lebesgue integral, but not too difficult.) Quick note about Charlie's answer: It is not a consequence of even or odd that the area under the curve is positive or negative, and positive + negative is not always zero. This is only true when it is x + (-x), where x is a real number (strictly finite). In fact, I can write an even function f(x) whose integral over any nondegenerate open interval is negative. Take for instance f(x) = -x^2 Also, sin(x) is odd because sin(-x) = -sin(x) cos(x) is even because cos(-x) = cos(x) These are just by the definitions of even and odd functions, and should not be associated with concepts of area, and definitely should not be associated with the sign in front of the respective areas to the left and right of the y-axis. This kind of explanation actually leads to more confusion than just saying "sin(x) is odd, and cos(x) is even", and then moving on to say that you can prove the desired results for f(x) = sin(x), f(x) = cos(x), and then trying to generalize this proof to general odd or even integrable functions. The problem with trying to generalize to odd or even integrable functions is that if we only know it for sin(x) and cos(x), we have to know a lot of Fourier analysis to be able to say anything about a general even f(x), or a general odd f(x). That being said, the direct proof above which uses nothing but properties of the integral, the definition of an integrable function, and the definition of even and odd functions.