Explanation
Solution
Hame integral ko do parts mein todna hoga: I=∫01xf(x)dx+∫12xf(x)dx.
Part 1: x∈[0,1) ke liye
Is range mein [x]=0 hota hai.
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f(x)=emin{x2,x−0}=emin{x2,x}
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Chunki 0 \le x < 1 hai, isliye x2≤x hamesha sahi hota hai.
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Isliye, f(x)=ex2.
Toh pehla integral:
Maana x2=t⟹2xdx=dt⟹xdx=2dt.
Limit badal kar 0 se 1 hi rahegi.
∫0121etdt=21[et]01=21(e−1)
Part 2: x∈[1,2) ke liye
Is range mein [x]=1 hota hai. Hame f(x)=e[x−logex] solve karna hai.
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Maana g(x)=x−logex.
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g′(x)=1−x1. Chunki x∈[1,2), g'(x) > 0 hai, matlab function badh raha hai (increasing).
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g(1)=1−loge1=1.
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g(2)=2−loge2≈2−0.693=1.307.
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Chunki g(x) ki value 1 aur 1.307 ke beech hai, iska Greatest Integer [g(x)] hamesha 1 hoga.
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Isliye, f(x)=e1=e.
Toh dusra integral:
∫12x⋅edx=e[2x2]12=e(24−21)=23e
Final Calculation:
Correct Option: (3) 2e−21
