Quod Erat Demonstrandum


[AL][PM] Definite integrals — trouble? (3)

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 12:45 上午

Here is the theorem about the validity of the method of substitution in evaluating definite integrals.Let y = f(x) be a continuous function on [a,b]. Suppose that x = g(t) is a function satisfying g(\alpha) = a, g(\beta) = b and g'(t) remains the same sign on (\alpha , \beta). Then

\int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

Let me copy the proof from the textbook directly as follows.

Suppose F(x) is a primitive function of f(x), i.e.

\frac{d}{dx}F(x) = f(x)

and so,

\frac{d}{dx}F[g(t)] = f[g(t)]g'(t)

Hence, F[g(t)] is a primitive function of f[g(t)]g'(t). Then,

\int_a^bf(x)dx = F(b) - F(a) = F[g(\beta)] - F[g(\alpha)] = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

On reading the proof above, I puzzled that there is nothing to say about the condition that “the sign of g'(t) remains the same over certain range" in the proof. But after looking into the proof carefully, the condition is actually used. Where? It is something about the following implication

x = g(t)\:\Longrightarrow\:f(x) = f(g(t))

What? The above is a must, isn’t it? No, man. We must add the condition about g'(t) (as one of the ways) to ensure its validity. Caution: if x = g(t), we cannot come up with the conclusion that f(x) = f[g(t)] always. F.6 boys, when we are discussing function, don’t forget to consider the domain. f(x) and f(g(t)) may have different domains and hence they may represent different functions, and we cannot say immediately that f(x) = f[g(t)]!

I’d like to discuss this in details. Let me start with a simple example.

Let f:\mathbb{R}\rightarrow\mathbb{R} such that

f(x) = 2x + 1

Now, we use “substitution", let x = g(t) = \sin(t). How about the newly formed composite function f[g(t)] (i.e. f[\sin(t)] = 2\sin(t) + 1), can we say that f(x) = f[g(t)] now? No, of course. f(x) is very free to assume any real real value, but f[g(t)] is restricted on its domain [-1,1] (if t \in\mathbb{R}) and f can only assume values in [-1,3].

Ok, let’s get moving to the situation of integration. Refer to Fig. 1 & 2 below

x = g(t) is assumed to be differentiable on (\alpha,\beta), then is it true to say \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt?

No! At least, f(x) is defined on [a,b] (see Fig. 1) while f[g(t)] is NOT defined for all t\in [\alpha,\beta], it is because, when t varies from \gamma to \beta, the value of x is greater than b (see Fig. 2) and f(x) is NOT defined for x > b (see Fig. 1 again). If the condition “g'(t) remains the same sign on (\alpha , \beta)" is added, as shown in Fig. 4 below,

it is certain that f(x) = f[g(t)] for t\in [\alpha,\beta] and hence \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt.

Fine? Discussion ends? Not yet. I’d to add more. When the condition “g'(t) remains the same sign on (\alpha , \beta)" is satisfied, the method of substitution is valid. How about when the condition is NOT satisfied, will the method of substitution be failed? Under basic training in simple logic, we CANNOT say so.

I think that the condition “g'(t) remains the same sign on (\alpha , \beta)" is too strong! To ensure the validity of the method of substitution, it is not necessary to have the condition above, let me suggest an example below (for my poor mathematical skills, I just use figures to illustrate without rigorous real analysis treatment, sorry.)

Fig. 6 shows the graph of x = g(t) which is assumed to be differentiable over (\alpha,\beta). Obviously, the so-called condition is not satisfied. However, in this case, we still have \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt, can you see why?

Note that, in Fig. 6, g'(\mu) = g'(\lambda) = 0, hence we have g'(x) does not change signs over the intervals (\alpha,\gamma), (\gamma,\mu) and (\mu,\beta) respectively and hence the validity of the method of substituion is confirmed; therefore

\int_a^df(x)dx = \int_{\alpha}^{\lambda}f[g(t)]g'(t)dt
\int_d^cf(x)dx = \int_{\lambda}^{\mu}f[g(t)]g'(t)dt
\int_c^bf(x)dx = \int_{\mu}^{\beta}f[g(t)]g'(t)dt

Summing up the three equations above, yielding

\int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

Yes, the method of substitution is valid even by using “ugly" differentiable functions like Fig. 7 below

The discussions above drove me to re-think about the condition “g'(t) remains the same sign on (\alpha , \beta)“, I think it is not quite necessary now; instead, I prefer using “g(\alpha) = a, g(\beta) = b, g(t) is differentiable on (\alpha, \beta) and g([\alpha,\beta]) = [a,b]“. Do you agree?

Nelson gave me a note from mathematical analysis used in university, and I found that the way of presentation of the method of substitution is very clear (and clever), let me present here to round up my utterance:

Thereom (Change of Variable Formula): If \phi : [a,b] \longrightarrow \mathbb{R} is differentiable, \phi' is integrable on [a,b] and f is continuous on \phi([a,b]), then \int_{\phi(a)}^{\phi(b)}f(t)dt = \int_{a}^{b}f[\phi(x)]\phi'(x)dx.

Proof: Let g(x) = \int_{\phi(a)}^{\phi(b)}f(t)dt. Then g'(t) = f[\phi(x)]\phi'(x). So \int_a^bf[\phi(x)]\phi'(x)dx = \int_a^bg'(x)dx = g(b) - g(a) = \int_{\phi(a)}^{\phi(b)}f(t)dt.


1 則迴響 »

  1. […] [AL][PM] Definite integrals — trouble? (3) […]

    通告 由 積分二三事 | Quod Erat Demonstrandum — 2014/06/16 @ 11:50 下午 | 回應

RSS feed for comments on this post. TrackBack URI



WordPress.com 標誌

您的留言將使用 WordPress.com 帳號。 登出 /  變更 )

Google+ photo

您的留言將使用 Google+ 帳號。 登出 /  變更 )

Twitter picture

您的留言將使用 Twitter 帳號。 登出 /  變更 )


您的留言將使用 Facebook 帳號。 登出 /  變更 )


連結到 %s

在 WordPress.com 建立免費網站或網誌.

%d 位部落客按了讚: