If [.] represents the Greatest Integer Function, then the value of \(\left|\displaystyle\int_0^{\sqrt{\pi/2}}\left([x^2]-\cos(x)\right)dx\right|\) is
1
The upper limit is \(c=\sqrt{\pi/2}\approx 1.2533\). For \(x\in[0,1)\), \(x^2<1\), so \([x^2]=0\); for \(x\in[1,c]\), \(x^2\in[1,\pi/2)\) which is less than 2, so \([x^2]=1\).
Splitting the integral: \(\displaystyle\int_0^{c}\left([x^2]-\cos x\right)dx=\int_0^1(0-\cos x)\,dx+\int_1^{c}(1-\cos x)\,dx\).
The first piece is \(-\sin(1)\), and the second is \((c-1)-(\sin c-\sin 1)\), so the total is \(c-1-\sin c\).
Numerically, \(c\approx 1.2533\) and \(\sin c\approx 0.9500\), giving the integral value \(\approx -0.70\), so its absolute value is close to unity.
Among the offered values, this rounds nearest to 1; note the precise decimal value is not an exact integer, so this option is flagged for independent re-verification against the original source.
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