A) \[\frac{5}{2}\]
B) \[5\]
C) \[\frac{69}{2}\]
D) \[\frac{71}{2}\]
Correct Answer: B
Solution :
| Let \[I=\int_{0}^{5}{[x]\{x\}dx}\] |
| \[\Rightarrow \] \[I=\int_{0}^{1}{\{0\}dx}+\int_{1}^{2}{\{x\}dx+2\int_{2}^{3}{\{x\}dx}}\]\[+\,3\int_{3}^{4}{\{x\}dx+\int_{4}^{5}{4\{x\}dx}}\] |
| \[\Rightarrow \] \[I=(1+2+3+4)\int_{0}^{1}{x\,dx}\] [\[\because \{x\}\]is periodic with 1] |
| \[\Rightarrow \] \[I=10\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}\] |
| \[I=\frac{10}{2}=5\] |
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