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JEE Mains · Maths · STD 12 - 7.2 definite integral

Let \(f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}\) where \([ t\) ] denotes the greatest integer \(\leq t\).Then\(\int_{0}^{10} f(x) d x+\int_{0}^{10}(f(x))^{2} d x+\int_{0}^{10}|f(x)| d x\) is equal to

  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

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