Question:
Consider the integral
\(I = \int_0^{10}\frac{[x]e^{[x]}}{e^{x-1}}\,dx\),
where \([x]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(I\) is equal to :
Consider the integral
\(I = \int_0^{10}\frac{[x]e^{[x]}}{e^{x-1}}\,dx\),
where \([x]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(I\) is equal to :
\(I = \int_0^{10}\frac{[x]e^{[x]}}{e^{x-1}}\,dx\),
where \([x]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(I\) is equal to :
Show Hint
Split the integral over unit intervals where \([x] = n\) is constant.
Updated On: Nov 4, 2026
- \(45\,(e - 1)\)
- \(45\,(e + 1)\)
- \(9\,(e + 1)\)
- \(9\,(e - 1)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Split \([0,10]\) into \([n, n+1)\), \(n = 0,\dots,9\), where \([x] = n\):
\(I = \sum_{n=0}^{9} n e^{n}\int_n^{n+1} e^{1-x}\,dx\).
\(\int_n^{n+1} e^{1-x}\,dx = e\left(e^{-n} - e^{-n-1}\right) = e^{-n}(e - 1)\).
So each term is \(n e^n\cdot e^{-n}(e-1) = n(e-1)\).
\(I = (e-1)\sum_{n=0}^{9} n = 45(e-1)\).
Answer: (1) \(45\,(e - 1)\)
\(I = \sum_{n=0}^{9} n e^{n}\int_n^{n+1} e^{1-x}\,dx\).
\(\int_n^{n+1} e^{1-x}\,dx = e\left(e^{-n} - e^{-n-1}\right) = e^{-n}(e - 1)\).
So each term is \(n e^n\cdot e^{-n}(e-1) = n(e-1)\).
\(I = (e-1)\sum_{n=0}^{9} n = 45(e-1)\).
Answer: (1) \(45\,(e - 1)\)
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