Consider the integral I=∫ _0^10 [x]e^[x] e^x-1dx where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :

$$I={\int }_{0}^{10}[x]\cdot {e}^{[x]-x+1}$$I=∫010​[x]⋅e[x]−x+1

$$I={\int }_{0}^{1}0dx+{\int }_{1}^{2}1\cdot {e}^{2-x}+{\int }_{2}^{3}2\cdot {e}^{3-x}+\ldots ..+{\int }_{9}^{10}9\cdot {e}^{10-x}dx$$I=∫01​0dx+∫12​1⋅e2−x+∫23​2⋅e3−x+…..+∫910​9⋅e10−xdx

$$\Rightarrow I=\sum _{n=0}^{9}{\int }_{n}^{n+1}n\cdot {e}^{n+1-x}dx$$⇒I=n=0∑9​∫nn+1​n⋅en+1−xdx

$$=-\sum _{n=0}^{9}n{({e}^{n+1-x})}_{n}^{n+1}$$=−n=0∑9​n(en+1−x)nn+1​

$$=-\sum _{n=0}^{9}n\cdot ({e}^{0}-{e}^{1})$$=−n=0∑9​n⋅(e0−e1)

$$=(e-1)\sum _{n=0}^{9}n$$=(e−1)n=0∑9​n

$$=(e-1)\cdot \frac{9\cdot 10}{2}$$=(e−1)⋅29⋅10​

$$=45(e-1)$$=45(e−1)