Apparent contradiction in integration by substitution

Question

Let $$I:=\frac{4}{3}=\int\limits_{-1}^12x^2\mathrm{d}x=\int\limits_{-1}^1x\cdot2x\mathrm dx$$ and $$\varphi:[-1,1]\to[0,1],\ x\mapsto x^2\implies\varphi'(x)=2x,\quad x=\sqrt{\varphi(x)}$$ so $I$ can be written as $$I=\int\limits_{-1}^1f(\varphi(x))\varphi'(x)\mathrm dx,\qquad f:[0,1]\to R,\ t\mapsto\sqrt t.$$ So $\varphi:[a,b]\to I$ is differentiable with a continuous derivative, $I\subseteq R$ is an interval and $f:I\to R$ is a continuous function (compare to the Wikipedia statement of the theorem). But then integration by substitution can be applied: $$I=\int\limits_{\varphi(-1)}^{\varphi(1)}f(t)\mathrm dt=\int\limits_1^1\sqrt t\mathrm dt=0,$$ a contradiction. So what did I do wrong?

Answer 1

Actually, $x\neq\sqrt{\varphi(x)}$ for all $x<0$ - for example let $x=-1$: $$\sqrt{\varphi(x)}=\sqrt{(-1)^2}=\sqrt{1}=1\neq-1$$ But the integration is from $-1$ to $1$, so $x$ has to be allowed to be negative. If one splits the integral though, like $$I=\int\limits_{-1}^02x^2\mathrm dx+\int\limits_0^12x^2\mathrm dx=2\int\limits_0^12x^2\mathrm dx,$$ the result of the integration by substitution is actually correct: $$I=2\int\limits_{\varphi(0)}^{\varphi(1)}f(t)\mathrm dt=2\int\limits_0^1\sqrt t\mathrm dt=2\left[\frac{2}{3}t^{\frac{3}{2}}\right]_0^1=\frac{4}{3}.$$ This took me way too long to realize.