Why can't $\int_0^1\sin(x^2) dx$ be equal to $2$?
What makes this true? Intuitively, it makes sense. But why?
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While Peter's answer is right, I think there's a more intuitive estimate. We know that $|\sin(y)| \leq 1$, and thus $\displaystyle \left|\int_0^1 \sin(x^2)dx \right| \leq \int_0^1 1 dx = 1$, the area of a square of side-length $1$.
Note that $$\sin(x^2)\leq x^2$$
$$\int_0^1\sin(x^2)<\int_0^1x^2=\frac 1 3$$
Even simpler, $0 \leq \sin \theta \leq 1$ for $\theta \in [0, 1] \subset [0,\pi]$, so $\int_0^1 \sin(x^2) dx \leq 1$.
Draw a rectangle with vertices at $(0,0), (1,0), (1,0), (1,1)$. The graph of $\sin(x^2)$ between 0 and 1 fits completely inside this rectangle, because $\sin(0)=0$ and $\sin(1)<1$.
The integral is the area of the part of the rectangle under the curve, and so must be less than the area of the entire rectangle, which is 1, which is less than 2.