Consider the integral I=∫010ex−1[x]e[x]dx where [x] denotes the greatest integer less than or equal to x. Then, the value of I is equal to
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Official solution
\[I=\int_0^{10}\frac{[x]e^{[x]}}{e^{x-1}}dx=e\int_0^{10}[x]e^{[x]}e^{-x}dx\]
Splitting into unit intervals [n,n+1) where [x]=n:
\[I=e\left[\int_0^1 0\,e^xdx+\int_1^2 e\cdot e^{-x}dx+\int_2^3 2e^2\cdot e^{-x}dx+\cdots+\int_9^{10}9e^9\cdot e^{-x}dx\right]\]
Using ∫aa+1e−xdx=e−a−e−a−1=e−a(1−e−1), each term nen⋅e−n(1−e−1)=n(e−1)⋅e−1⋅e... simplifying each bracket term gives n(e−1) for n=1,…,9:
\[I=e\cdot\frac{1}{e}\sum_{n=1}^{9}n(e-1)=(1+2+\cdots+9)(e-1)=\frac{9\times10}{2}(e-1)=45(e-1)\]