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Suppose that $a>0$ and that $f$ is integrable on $[-a,a]$. Show that if $f$ is even then $$ \int_{-a}^0 fdx = \int_0^a fdx $$ using the Riemann sum definition of Riemann integrability.

This is what I have tried:

Let $P$ be tagged partition of $[-a,0]$ i.e. $-a = x_0 < x_1 < \ldots < x_k = 0$ and $P^*$ be the symmetrical tagged partition of $[0,a]$ i.e. $0 = x_k < x_{k+1} < \ldots < x_{2k} = a$ where $x_0=-x_{2k}, x_1=-x_{2k-1}, \ldots , x_{k-1}=-x_{k+1}$

Then I try to prove that $S(f;P)=S(f;P^*)$.

Define $S(f;P)= \sum_{i=1}^k f(x_i-1)\cdot(x_i - x_{i-1})$ and $S(f;P^*)= \sum_{i=k}^{2k} f(x_i)\cdot(x_i - x_{i-1})$. Since $f$ is even i.e. $f(x_{2k})=f(-x_0)=f(x_0)$ etc, $S(f;P)=S(f;P^*)$.

Does this proof look okay? Did I use and justify the definition correctly here? Can I do similar proof when f is odd?

Thanks.

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1 Answer 1

What you have so far is correct, but it is not a complete solution. You need to prove a statement about integrals, and what you have so far is a statement about sums. The transition from sums to integrals is the most delicate part of proofs in Riemann integration.

Details may depend on how the Riemann integral was developed in your course. For one thing, you should quote a result saying that if $f$ is integrable on $[-a,a]$, then it is integrable on subintervals, such as $[-a,0]$ and $[0,a]$.

Then it can go like this: for every $\epsilon>0$ there exists $\delta_1$ such that for every $\delta_1$-fine tagged partition $P$ on $[0,a]$ (that is, a partition with subintervals of length at most $\delta_1$) we have $$ \left|\int_0^a f(x)\,dx - S(f,P)\right|<\epsilon \tag1$$ Similarly, there exists $\delta_2$ such that for every $\delta_2$-fine tagged partition $P$ on $[-a,0]$ we have $$ \left|\int_{-a}^0 f(x)\,dx - S(f,P)\right|<\epsilon \tag2$$ Now take a $\delta$-fine partition $P$ of $[0,a]$ with $\delta=\min(\delta_1,\delta_2)$. Let $P^*$ be its reflection. Apply (1) to $P$ and (2) to $P^*$. Conclude that $$ \left|\int_{-a}^0 f(x)\,dx - \int_0^a f(x)\,dx\right|<2\epsilon \tag3$$ Make the final step based on $\epsilon$ being arbitrarily small.

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