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I am working on the following question:

Let $f$ be an integrable odd function, i.e. $f(-x)=-f(x)$ for all $x\in \mathbb{R}$. Show algebraically that $\int_{-a}^{a}f(x)dx=0$ for all real numbers $a$. Explain this fact geometrically.

I was wondering if I have the right idea with what I have below.

Let $I=\int_{-a}^{a}f(x)dx$.

$$\int_{-a}^{a}f(x)dx=\int_{-a}^{0}f(x)dx+\int_{0}^{a}f(x)dx$$

Let $A_{1}=\int_{-a}^{0}f(x)dx$ and $A_{2}=\int_{0}^{a}f(x)dx$. Then, using the definition of an integral,

$$A_{2}=\lim_{n \to \infty }\sum_{i=1}^{n}f(x_{i})\Delta x=\lim_{n \to \infty }\sum_{i=1}^{n}f(x_{i})\frac{a}{n}=\frac{a}{n}\lim_{n \to \infty }\sum_{i=1}^{n}f(x_{i})$$

Where $\Delta x=\frac{a}{n}$. For the other area,

$$A_{1}=\lim_{n \to \infty }\sum_{i=1}^{n}f(-x_{i})\Delta x =\lim_{n \to \infty }\sum_{i=1}^{n}f(-x_{i})\frac{a}{n}$$$$=\frac{a}{n}\lim_{n \to \infty }\sum_{i=1}^{n}f(-x_{i})=\frac{a}{n}\lim_{n \to \infty }\sum_{i=1}^{n}-f(x_{i})=\frac{-a}{n}\lim_{n \to \infty }\sum_{i=1}^{n}f(x_{i})=-A_{2}$$

From above, $I=A_{1}+A_{2}=(-A_{2})+A_{2}=0$

I feel this wont get full marks.

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3  
Use the substitution $u = -x$ –  kahen Jan 18 '12 at 20:39
1  
Yes, no need to go into Riemann sums. Make the substitution just suggested, u = -x, in the first integral, A1, and then show it is equal to the negative of A2. –  Simon S Jan 18 '12 at 20:45
    
Of course, that's assuming you've gotten to that point... –  David Mitra Jan 18 '12 at 20:49

1 Answer 1

up vote 2 down vote accepted

If you don't need to go all the way to the definition via Riemann sums:

After you show (or already know) that $f$ being integrable on $[-a,a]$ implies being integrable separately on $[-a,0]$ and on $[0,a]$, you can use the change of variable $u=-x$ on $\int_{-a}^0 f(x)\,dx$ to get $$\int_{-a}^0 f(x)\,dx = \int_{a}^{0}f(-u)(-\,du) = -\int_a^0(-f(u))\,du = \int_a^0f(u)\,du = -\int_0^af(u)\,du = -\int_0^af(x)\,dx.$$ And then it easily gives $$\int_{-a}^af(x)\,dx = \int_{-a}^0f(x)\,dx + \int_0^af(x)\,dx = -\int_0^af(x)\,dx + \int_0^af(x)\,dx = 0.$$


If you need to use the definition via Riemann sums:

It is generally okay; but one possible problem is that you should explain why the fact that $f$ is integrable on $[-a,a]$ implies that it is separably integrable on $[-a,0]$ and on $[0,a]$. Another issue might be how you are writing the Riemann sums for $A_1$; it would be best to begin with a partition on $[-a,0]$ and then explain how to transform it into a Riemann sums on $[0,a]$.

You can take care of all of this in one step by considering instead Riemann sums and partitions on $[-a,a]$ directly: for each $n$, let $P_n$ be the partition of $[-a,a]$ into $2^n$ equal parts: this will guarantee that the partition is symmetric about the origin. That is, if we let $\Delta x = \frac{2a}{2^n}$, and set $$-a = x_0 \lt x_1 \lt x_2 \lt\cdots \lt x_{2^n-1} \lt x_{2^n} = a,$$ with $x_{i+1}-x_i = \Delta x$, then $x_{i} = -x_{2^n-i}$ for $i=0,1,\ldots,2^{n-1}$. (In particular, $x_{2^{n-1}} = -x_{2^{n-1}} = 0$).

Then select $y_i \in [x_i,x_{i+1}]$ to be the midpoint of the interval for $i=0,\ldots,2^{n}-1$. Then we have $y_i=-y_{2^n-i-1}$. So the $n$th Riemann sum is: $$\begin{align*} S_n &= \sum_{i=0}^{2^n-1}f(y_i)\Delta x = \sum_{i=0}^{2^{n-1}-1}f(y_i)\Delta x + \sum_{i=2^{n-1}}^{2^n-1}f(y_i)\Delta x\\ &= \frac{2a}{n}\sum_{i=0}^{2^{n-1}-1}f(y_i) + \frac{2a}{n}\sum_{i=2^{n-1}}^{2^n-1}f(y_i)\\ &=\frac{2a}{n}\left( \sum_{i=0}^{2^{n-1}-1}f(-y_{2^n-i-1}) + \sum_{i=2^{n-1}}^{2^n-1}f(y_i)\right)\\ &= \frac{2a}{n}\left( -\sum_{i=0}^{2^{n-1}-1}f(y_{2^n-i-1}) + \sum_{i=2^{n-1}}^{2^n-1}f(y_i)\right) \end{align*}$$ Now do a change of index in the first sum, by letting $j=2^n-i-1$; the sum will then range from $j=2^n-1$ down to $2^n-(2^{n-1}-1)-1 = 2^{n-1}$, and you can do the sum in the opposite order. Then things will cancel out nicely and you will get that the Riemann sum is always equal to $0$.

Since the function is Riemann integrable, any choice of partitions and points (that is, any choice of Riemann sums) will have the value of the integral as a limit, and you are done.

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