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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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Forgive me if I'm missing something, but why would it be 2? I mean, with so many numbers to choose from, why would you expect it to be 2? – Javier Aug 13 '12 at 3:03
What on Earth would make you think that $\int_0^1 \sin(x^2)dx = 2$?! – user5137 Aug 13 '12 at 3:09
Joke: $(1-0)*\text{The 2 in }x^2 = 2$ – Frenzy Li Aug 13 '12 at 4:12
@FrenzYDT. I don't get it. – Pedro Tamaroff Aug 13 '12 at 4:52
$\sin$ is too small. – André Nicolas Aug 13 '12 at 5:53
up vote 21 down vote accepted

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 $1$.

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Bingo........... – copper.hat Aug 13 '12 at 1:57

Note that $$\sin(x^2)\leq x^2$$

This gives

$$\int_0^1\sin(x^2)<\int_0^1x^2=\frac 1 3$$

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Even simpler, $0 \leq \sin \theta \leq 1$ for $\theta \in [0, 1] \subset [0,\pi]$, so $\int_0^1 \sin(x^2) dx \leq 1$.

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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 $\sin(1)<1$.

graph of sin(x^2)

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.

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