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JEE Mains · Maths · STD 12 - 7.2 definite integral

यदि समाकलन \(\int \limits_{0}^{5} \frac{ x +[ x ]}{ e ^{ x -[ x ]}} dx =\alpha e ^{-1}+\beta\) है, जहाँ \(\alpha, \beta \in R , 5 \alpha+6 \beta=0\) तथा \([ x ]\) महत्तम पूर्णांक \(\leq x\) है, तो \((\alpha+\beta)^{2}\) का मान बराबर है -

  1. A \(100\)
  2. B \(25\)
  3. C \(16\)
  4. D \(36\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(25\)

Step-by-step Solution

Detailed explanation

\(I=\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,d x\) \(\int_{0}^{1} \frac{x}{e^{x}} d x+\int_{1}^{2} \frac{x+1}{e^{x-1}} d x+\int_{2}^{3} \frac{x+2}{e^{x-2}} d x+\ldots . \int_{4}^{5} \frac{x+4}{e^{x-4}} \,d x\) \(\quad\quad\quad\quad\quad x=t+1\quad x=z+2 \quad\quad\quad x=y+4\)…
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