$u$-substitution always evaluates to $0$

Question

Consider the integral

$$ \int_a^b f(x) \,dx $$

now make the $u$-substitution $u \mapsto c + (x-a)(x-b)$. The resulting integral is

$$ \int_c^c h(u) \,du $$

where $h(u)$ is the integrand $f$ after the substitution, however, regardless of $f$ the integral $\int_c^c du = 0$. Looking at the definition Wikipedia provides I believe the substitution meets every condition. It's differentiable and has a integrable derivative, because it's a polynomial. So what's wrong with this substitution such that it always results in $0$?

Answer 1

The formula $$\int_{g(a)}^{g(b)} f(x) \, dx=\int_{a} ^{b} f(g(t)) g'(t) \, dt$$ holds without $g$ being bijective. But as you can see the substitution has to be of the form $x=g(t) $ and you are trying to make a substitution using $t=h(x) $ and this would require that $h$ is invertible to put it in the form $x=h^{-1}(t)$ so that the above Rule for change of variables applies. Your substitution does not have this property and hence the conclusion obtained is wrong.

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It is better to understand the theorems completely before applying them. And more often than not most people remember only the conclusions of the theorems rather than their hypotheses. By habit one should avoid this and understand how the hypotheses lead to conclusions and what happens when one of the hypotheses is not satisfied.

Answer 2

The formula of change of variables is, for $\phi\in\mathcal{C}^1$, $$\int_{\phi(a)}^{\phi(b)}f(t)\,dt=\int_a^bf(\phi(t))\phi'(t)\,dt,$$ so to get what you write, you should have that $f$ is written as a function of the form $$\tilde{f}\circ \text{“your function''}\times\text{“the derivative of your function''},$$ or to inverse "your function", which is of course not possible since it is not bijective.

EDIT : More precisely, if you want to change both the integrand and the boundaries, you have to find $\tilde{a}\in\phi^{-1}(a)$ and $\tilde{b}\in\phi^{-1}(b)$ to finally write $$\int_a^bf(t)\,dt=\int_\tilde{a}^\tilde{b}f(\phi(t))\phi'(t)\,dt,$$ and if $a\neq b$ then $\tilde{a}\neq\tilde{b}$ (else $a=\phi(\tilde{a})=\phi(\tilde{b})=b$), and you can't conclude a priori that the right integral is $0$ using the same boundaries argument.

Answer 3

$$ \int_a^b f(x) \,\mathrm dx $$ now make the $u$-substitution $$u \mapsto c + (x-a)(x-b).$$

Typo: judging from your post, your substitution is actually \begin{align}u=g(x)&=c + (x-a)(x-b)\tag# \\ g'(x)&=2x-a-b\\g(a)&=c=g(b).\end{align}

The resulting integral is $$ \int_c^c h(u) \,\mathrm du =0.$$

what's wrong with this substitution such that, regardless of the starting integrand $f,$ it results in $0\ ?$

It is indeed correct that $$ \int_a^b f(x) \,dx\overset{u = c + (x-a)(x-b)}{=}0.\tag✓$$ However, this is not “regardless of $f,$” but whenever the substitution is even possible, which is precisely when $f$ has rotational symmetry about $\left(\frac{a+b}2,0\right)$ and $h$ is integrable, and has an antiderivative, on $g[a,b].$ (The stipulation on $h$ means that the red bit in $$ \int_c^c \color{red}{\text{[this integrand doesn't matter]}} \,dx=0$$ is false.) So, if $f$ lacks the stipulated symmetry property, then there is no basis for, and no way to, make the given substitution $(\#).$

Proof

Let's begin with this strong version of the integration-by-substitution theorem, which does not generally require injectivity:

• If $g'$ is integrable on $[a,b]$ and $h$ is integrable, and has an antiderivative, on $g[a,b],$ then $$\int_a^bh\big(\color{violet}{g(x)}\big)\,\color{cyan}{g'(x)}\,\mathrm{d}x=\int_{g(a)}^{g(b)}h(u)\,\mathrm{d}u.$$

This corresponds exactly to your above work, with $$f(x)=h\Big(\color{violet}{c + (x-a)(x-b)}\Big)\,\color{cyan}{(2x-a-b)}.$$ As $f$ has not been concretely given, function $h$ is necessarily abstract (though we note that the choice of $h$ is also to satisfy the theorem's conditions; this immediately happens if $h$ is continuous on $g[a,b].)$ Now let's shift $f$ by $\displaystyle\frac{a+b}2$ units leftwards: $$f_N(x)=f\left(x+\frac{a+b}2\right)\\ =h\Bigg(c + \left(\left(x+\frac{a+b}2\right)-a\right)\left(\left(x+\frac{a+b}2\right)-b\right)\Bigg)\:\:\left(2\left(x+\frac{a+b}2\right)-a-b\right)\\ =h\left(c+x^2-\frac{(a-b)^2}4\right)\;\;2x.$$ Since $f_N(-x)=-f_N(x),$ the new function is odd. Thus, $f$ has rotational symmetry about the point $\left(\frac{a+b}2,0\right).$

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Addendum

In the youtube video Weird Integral Trick, @blackpenredpen demonstrates that $$u=g(x)=x(x-a-b)$$ is a wonderful substitution, because it amazingly and universally zeroes out every integral: $$\int_a^b f(x)\,\mathrm dx\overset{u=x(x-a-b)}{=}\int_{-ab}^{-ab}\text{[this integrand doesn't matter]}\,\mathrm du=0.$$

The apparent anomaly is currently explained in neither the video nor its comments, so, for the record, it has the same explanation as above: this substitution is possible precisely when both these conditions are satisfied: 1) $f$ on $[a,b]$ has rotational symmetry about $\left(\frac{a+b}2,0\right),$ 2) $h$ on $g[a,b]$ is integrable and has an antiderivative; if so, then the result $0$ is correct.

Answer 4

Assume $a<b$.

If $u=c+(x-a)(x-b)$. Then

\begin{align} u&=x^2-(a+b)x+ab+c\\ &=\left(x-\frac{a+b}{2}\right)^2+ab-\left(\frac{a+b}{2}\right)^2+c\\ &=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2+c\\ x&=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2-c-u} \end{align}

When $\displaystyle x<\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and

$$\frac{du}{dx}=2x-(a+b)=-\sqrt{(a-b)^2-4c-4u}$$

When $\displaystyle x\ge\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and

$$\frac{du}{dx}=2x-(a+b)=\sqrt{(a-b)^2-4c-4u}$$

When $\displaystyle x=\frac{a+b}{2}$, $\displaystyle u=c-\left(\frac{a-b}{2}\right)^2$

Therefore,

\begin{align} \int_a^bf(x)dx&=\int_a^{\frac{a+b}{2}}f(x)dx+\int_{\frac{a+b}{2}}^bf(x)dx\\ &=\int_0^{c-\left(\frac{a-b}{2}\right)^2}\left[\frac{f\left(\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{-\sqrt{(a-b)^2-4c-4u}}\right]du\\ &\qquad +\int_{c-\left(\frac{a-b}{2}\right)^2}^0\left[\frac{f\left(\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{\sqrt{(a-b)^2-4c-4u}}\right]du \end{align}

It looks quite complicated. But my point is that the definite integral should be broken into two parts with different integrands.

Answer 5

\begin{align} u & = c + (x-a)(x-b) \\[10pt] &= c + \left( x - \frac{a+b} 2 \right)^2 + \left( \frac{a-b} 2 \right)^2 & & \text{(completing the square)} \\[12pt] & \text{When } x = a \text{ or } x=b \text{ then } u = c. \\ & \text{When } x = \frac{a+b}2 \text{ then } u = c+ \left( \frac{a+b} 2 \right)^2. \end{align} $$ \int_a^b f(x)\,dx = \int_a^{(a+b)/2} f(x) \,dx + \int_{(a+b)/2}^c f(x)\, dx\quad\longleftarrow \text{two terms} $$ $$ du = 2\left( x - \frac{a+b} 2 \right) \, dx = \Big( \text{some function of } u \Big)\, dx $$ This "some function" is one thing in one of the "two terms" above and is another thing in the other one of the "two terms" above. The reason is that when you find $x-\dfrac{a+b}2$ as a function of $u,$ you get the usual “plus-or-minus” $(\text{“}\pm\text{''})$ issue that you see when solving quadratic equations. In the first of the "two terms" above, you need the solution with the minus sign, and in the second the one with the plus sign.

Answer 6

Your error is thinking that "the integrand $f$ after substitution" is well defined. (actually, the integrand is $f(x) \mathrm{d} x$, but that distinction is not relevant to the topic)

It is indeed true that if $f$ is a function with the property that there exists a function $h$ such that

$$ f(x) \mathrm{d} x = h(u) \mathrm{d}u $$

(where $u$ and $x$ are related by $u = c + (x-a)(x-b)$), then you do have

$$ \int_a^b f(x) \mathrm{d}x = \int_c^c h(u) \mathrm{d}u = 0 $$

However, you are not guaranteed such an $h$ exists. Instead, the generally $2$ to $1$ nature of the relationship between $x$ and $u$ means that on any path where $x$ varies from $a$ to $b$, you will get one function $h_0$ on part of the path and another function $h_1$ on the rest of the path. Then, the best you can say is that

$$ \int_a^b f(x) \, \mathrm{d}x = \int_c^d h_0(u) \mathrm{d}u + \int_d^c h_1(u) \mathrm{d}u $$