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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(f ( x )=\min \{[ x -1],[ x -2], \ldots,[ x -10]\}\) है, जहाँ [t] महत्तम पूर्णांक \(\leq t\) है। तब \(\int_0^{10} f ( x ) dx +\int_0^{10}( f ( x ))^2 dx +\int_0^{10}| f ( x )| dx\) बराबर है  \(..........\)

  1. A \(384\)
  2. B \(385\)
  3. C \(386\)
  4. D \(387\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(385\)

Step-by-step Solution

Detailed explanation

\(f(x)=[x]-10\) \(\int_{0}^{10} f(x) \cdot d x=-10-9-8-\ldots . .-1\) \(=-\frac{10 \cdot 11}{2}=-55\) \(\int_{0}^{10}(f(x))^{2} d x=10^{2}+9^{2}+8^{2}+\ldots+1^{2}\) \(=\frac{10 \cdot 11 \cdot 21}{6}=385\) \(\int_{0}^{10}|f(x)|=10+9+8+\ldots .+1\) \(=\frac{10 \cdot 11}{2}=55\)…
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