Answer
$\int_0^{1} \sin (x^2) dx$ cannot possibly equal to $2$
Work Step by Step
Since, we have $-1 \leq \sin (x^2) \leq 1$ for all $x \in [0,1]$
Now, we have $ \int_0^{1} f(x) dx \leq \int_0^{1} g(x) dx \leq \int_0^{1} h(x) dx$
But $ \int_0^{1} f(x) dx =-1(1-0)=-1$
and
$ \int_0^{1} h(x) dx=1(1-0)=1$
Then, we get $-1 \leq \int_0^{1} \sin (x^2) dx\leq 1$
Therefore, $\int_0^{1} \sin (x^2) dx$ cannot possibly equal $2$
