Every Integral equals to zero?

Question

$$\int_a^b{f(x)}dx$$ let $u=k(x),$ where $ k(a)=0, k(b)=0 \\ k^{-1}(u)=x \\ (k^{-1})'(u) du =dx$

$$\int_{k(a)}^{k(b)} f(k^{-1}(u))\times (k^{-1})'(u) du$$ $$=\int_{0}^{0} f(k^{-1}(u))\times (k^{-1})'(u) du$$ $$=0$$

If this is correct, then every integral would equal to zero, thus it has to be incorrect.

You'll have hard time finding an invertible map $k$ which has two inverse images for $0$. — mathcounterexamples.net, Oct 29 '19 at 11:06
Well, since $k(a)=k(b),$ $k$ can't be injective, and thus, $k^{-1}$ isn't actually well-defined. — WoolierThanThou, Oct 29 '19 at 11:06

score 3 · Answer 1 · answered Oct 29 '19 at 11:06

3

If $k(a) = 0 = k(b)$ then there is not exist $k^{-1}$

Well this $k$ is not inyective

answered Oct 29 '19 at 11:06

ZAF

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score 2 · Answer 2 · answered Oct 29 '19 at 11:22

Let us play the game with $$k(x):=(x-a)(b-x)$$ so that

$$dk=(a+b-2x)\,dx.$$

When $x$ moves from $a$ to $b$, $k$ increases from $0$ to $\dfrac{(b-a)^2}4$, then decreases back to $0$. At the same time, the sign of $dk$ changes from positive to negative.

Hence we will have a partial integral from $a$ to $\dfrac{a+b}2$ with a positive sign and another from $\dfrac{a+b}2$ to $0$ with a negative sign. Hence these two integrals do not cancel each other.

Every Integral equals to zero?

2 Answers2