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JEE Mains · Maths · STD 12 - 7.2 definite integral

Consider the integral \(I=\int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} d x,\) where \([ x ]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(I\) is equal to:

  1. A \(9( e -1)\)
  2. B \(45( e +1)\)
  3. C \(45( e -1)\)
  4. D \(9( e +1)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(45( e -1)\)

Step-by-step Solution

Detailed explanation

\(I =\int_{0}^{10}[ x ] \cdot e ^{[ x ]- x +1}\) \(I =\int_{0}^{1} 0 d x +\int_{1}^{2} 1 \cdot e ^{2- x }+\int_{2}^{3} 2 \cdot e ^{3- x }+\ldots . .+\int_{9}^{10} 9 \cdot e ^{10- x } dx\) \(\Rightarrow I=\sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n+1-x} d x\)…