Consider the integral ∫^0_10 [x]e^[x]/e^(x-1) dx,

Question

Consider the integral

\(\int\limits_{0}^{10}\frac{[x]e^{[x]}}{e^{x-1}}dx,\)

where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to:

(1) 9(e - 1)

(2) 45(e + 1)

(3) 45(e - 1)

(4) 9(e + 1)

Answer 1

Correct answer is (3)