There exists $c \in [0,1]$ for which $\int_{0}^{1}\sin(x^3) = \int_{0}^{c}\sin(x^2)$ T/F: There exists $c \in [0,1]$ for which $\int_{0}^{1}\sin(x^3) = \int_{0}^{c}\sin(x^2)$

I know the answer it true, and I already saw the proof. What I don't get is this:

$\sin$ is monotonically increasing in $[0, 1]$, so from the monotonocity of integrals:

$$\sin(x^3) \leq \sin(x^2) \quad \forall x \in [0, 1]$$
$$\int_{0}^{1}\sin(x^3) \leq \int_{0}^{1}\sin(x^2) \quad \forall x \in [0, 1]$$
$$L = \int_{0}^{1}\sin(x^3) \leq \int_{0}^{c}\sin(x^2) + \int_{c}^{1}\sin(x^2) = R \quad \forall x \in [0, 1]$$

So if there exists such $c$, it means that:

case 1: $L &lt; R \implies \int_{c}^{1}\sin(x^2) &lt; 0 \quad$  which is false.
case 2: $L = R \implies \int_{c}^{1}\sin(x^2) = 0 \implies c = 1 \implies \int_{0}^{1}\sin(x^3) = \int_{0}^{1}\sin(x^2) \quad$ which is again, false

So what's going on?
Let $L=\int_0^1\sin(x^3)\,dx$ and consider
$$
f(t)=\int_0^t \sin(x^2)\,dx
$$
Then $f(0)\le L\le f(1)$ and by the intermediate value theorem $L=f(c)$ for some $c\in[0,1]$.

About your doubts. It doesn't make sense to say
$$
\int_0^1\sin(x^3)\,dx\le\int_0^1\sin(x^2)\,dx
\qquad \text{for all $x\in[0,1]$}
$$
as the integrals don't depend on $x$.

Then, indeed,
$$
L\le\int_0^c\sin(x^2)\,dx+\int_c^1\sin(x^2)\,dx
$$
just says that
$$
\int_c^1\sin(x^2)\,dx\ge0
$$
Hint : put $f(t)=\int_0^t \sin(x^2)dx$ to conclude