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$f:[a,b]\rightarrow R$ that is integrable on [a,b]

So we need to prove:

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

1.) So we'll use a property of definite integrals: (homogeny I think it's called?)

$$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$

2.) Great, now using the fundamental theorem of calculus:

$$-1\int_{-b}^{-a}f(x)dx=(-1)^2\int_{-a}^{-b}f(x)dx=\int_{-a}^{-b}f(x)dx$$

This is where I'm stuck. For some reason I think it might be smarter to skip step 2, to leave it asL

$$-1\int_{-b}^{-a}f(x)dx$$

because graphically, we've "flipped" the graph about the x-axis, but we're still calculating the same area. Proving that using properties seems to have stumped me.

I prefer hints over solutions, thanks.

RonaldB
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4 Answers4

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Your first step is mistaken: it seems that you mistake $\int _a ^b (-f) (x) \ \Bbb d x$ for $\int _a ^b f (-x) \ \Bbb d x$; these two are completely different, and the homogeneity property applies only to the first formula, not to the second.

Just use the substitution $y = -x$, this will solve the problem in no time.

Alex M.
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  • Thanks. Seems the late night got to me and I didn't quite fully comprehend the properties sheet given to us. I'll try this out.

    Edit: yep also solved it in about 20 seconds. It seems to me that being an engineer in a math course makes me overthink things way too often.

     – RonaldB Jun 10 '16 at 20:37
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Let f(x) = x. then f(-x) = -x. Substituting -a and -b in the limits of the integration will lead to be f(a) = a then f(-a) = -(-a) = a.

it's simply this that you are multiplying the limits and the function by -1. if both are multiplied then they would get neutralized,

Dinesh.hmn
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Your step 1 is wrong and you can realize the error by considering $a=0$, $b=1$, $f(x)=e^x$.

Then $$ \int_{0}^{1}e^x\,dx=e-1, \qquad -\int_{-1}^0e^x\,dx=\frac{1}{e}-1 $$ which are quite different.

You can prove the statement by the definition, I'll use Riemann sums. A Riemann sum for $\int_{a}^{b}f(x)\,dx$ consists first in a choice $S$ of points $$ a=x_0<x_1<x_2<\dots<x_{n-1}<x_n=b, \qquad c_i\in[x_{i-1},x_i],\ i=1,2,\dots,n $$ and in considering $$ \sigma(f;S)=\sum_{i=1}^n f(c_i)(x_i-x_{i-1}) $$ Define $\delta(S)=\max\{x_1-x_0,x_2-x_1,\dots,x_n-x_{n-1}\}$; then it's not much difficult to give a meaning to $$ \lim_{\delta(S)\to0}\sigma(f;S) $$ and, if this exists, it is called the integral.

Now note that for each Riemann sum for $f(x)$ over $[a,b]$ we can define a Riemann sum $\hat{S}$ for $g(x)=f(-x)$ over $[-b,-a]$ by simply taking the negative of each point (and renaming indices, if you prefer to make your life difficult). Then $$ \sigma(g;\hat{S})=\sum_{i=1}^n g(-c_i)(-x_{i-1}-(-x_i)) = \sum_{i=1}^n f(c_i)(x_i-x_{i-1}) = \sigma(f;S) $$ Thus the two limits are equal, because also each Riemann sum for $g$ over $[-b,-a]$ corresponds to a Riemann sum for $f$ over $[a,b]$, by the same construction.

Similarly if you define integrals with upper and lower sums.

If the function $f$ is continuous, you can use substitutions (through the fundamental theorem of calculus): \begin{align} \int_{-b}^{-a}f(-x)\,dx &=\int_{b}^{a}f(t)\cdot(-1)\,dt && -x=t,\quad dx=-dt \\ &=-\int_{b}^{a}f(t)\,dt \\ &=\int_{a}^{b}f(t)\,dt \\ &=\int_{a}^{b}f(x)\,dx && x=t,\quad dt=dx \end{align}

egreg
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  • The use of Riemann sums is the right approach here as we are only given the fact that $f$ is Riemann integrable here. +1 – Paramanand Singh Jun 11 '16 at 07:02
  • @ParamanandSingh: No, the "change of variable" in the Riemann integral is in fact a purely measure-theoretic result (related to the concept push-forward) combined with the differentiability of a diffeomorphism. The only requirement on $f$ is integrability; continuity is assumed in order to provide a simple proof, based upon the fundamental theorem of calculus, but in reality is conceptually not needed. – Alex M. Jun 11 '16 at 10:01
  • @ParamanandSingh: See also http://math.stackexchange.com/questions/340180/change-of-variables-formula-for-riemann-and-lebesgue-integration. – Alex M. Jun 11 '16 at 10:01
  • @AlexM. Of course one can generalize this change of variable, but I don't think it is necessary to develop the whole machinery for this case where we can simply apply the definition. – egreg Jun 11 '16 at 10:03
  • @egreg: The definition of what? The change of variable is not a definition, it is a theorem. And if we are interested here in applying it, not in proving it, why not use its most general form (which, in fact, is conceptually more elementary than yours because it does not need to use the topological structure)? – Alex M. Jun 11 '16 at 10:17
  • @AlexM. I agree that presenting change of variable for Riemann integral is very important; this case, easily derived from the pure definition of the integral, can be a prologue to the more interesting case. No big theory: just apply the definition. – egreg Jun 11 '16 at 10:30
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    @AlexM.: I did not understand the stuff mentioned in your comment because I don't have much idea of measure theory. I have interacted with you in past and I think you don't like the policy of "simpler is better". If a problem can be solved by simpler and more economic methods I prefer that. Moreover preference of simplicity is not an excuse to avoid any study of advanced theories/techniques. One should study/know advanced stuff if one can but one should use it for problems which necessitate its use and preferably avoid "sledgehammer" approach. – Paramanand Singh Jun 11 '16 at 11:13
  • @ParamanandSingh: ...And, remarkably, requiring $f$ to be continuous is more complicated than requiring it to merely integrable, so in order to achieve simplicity the continuity of $f$ should be replaced by integrability. I, too, aim for simplicity (over complexity), so I choose my hypotheses to achieve it. – Alex M. Jun 11 '16 at 15:16
  • @AlexM.: I am really unable to understand what your point is! Both the question and this answer only focus on Riemann integrable functions. The part about continuity is nowhere emphasized (except for your comment). BTW I disagree that continuous functions are somehow more complicated than functions which are continuous almost everywhere. In fact most of the development of real analysis (leading to measure theory) happened by studying more and more weird sorts of discontinuous functions and each time more complicated functions were constructed. – Paramanand Singh Jun 11 '16 at 19:07
  • @ParamanandSingh: Changes of variable can be performed for integrable only functions, while egreg assumes $f$ to be continuous in order to do that (in the second part of the answer). Accepting my remark would shorten the proof (eliminating the first part of the answer) and simplify its conceptual foundations (currently, there are two topological structures here: one of topological space, and one of measurable space with a measure; the general theorem for changes of variables gets rid of the former - isn't this better if you enjoy minimality?). Let's end it here, anyway. – Alex M. Jun 11 '16 at 21:34
  • @egreg. How do you change by the given substitution the integral bounds, so that instead of $-b$ you get $b$ based on the substitution $y = -x$ ? Thanks. – user996159 Feb 16 '23 at 21:05
  • @user996159 The substitution is $t=-x$; when $x=-a$, we have $t=a$ and similarly for the other bound. – egreg Feb 16 '23 at 22:31
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There is a small error in your first step, you didn't change the sign of the limits. Always keep in mind that after making any substitution, don't forget to change the limits. Since you have made a substitution $x=-t$, so change the limits accordingly. So corrected step 1 is $$-\int_{b}^af(x)dx$$ Now you can reverse the limits to get the correct answer i.e. $$\int_{a}^bf(x)dx$$

Vineet Mangal
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  • The OP hasn't performed any substitution. – Alex M. Jun 11 '16 at 09:52
  • I thought it was a substitution. But then what is the homogenity property ? Step 1 is incorrect. – Vineet Mangal Jun 11 '16 at 10:29
  • The homogeneity property says that $\int (-f) = - \int f$. I believe that this is what the OP has applied (which, of course, is not the right approach for the problem). The OP has probably mistaken $\int f(-x)$ for $\int -f(x)$ (which is not correct unless $f$ is linear - which is needlessly restrictive). – Alex M. Jun 11 '16 at 10:46