A) \[10{{\log }_{e}}(10e)\]
B) \[\frac{10e-1}{{{\log }_{e}}10e}\]
C) \[\frac{10e}{{{\log }_{e}}10e}\]
D) \[(10e){{\log }_{e}}(10e)\]
E) \[3\,sq\,unit\]
Correct Answer: B
Solution :
Let \[I=\int_{1}^{e}{{{10}^{{{\log }_{e}}x}}dx}\] Again, let \[{{I}_{1}}=\int{{{10}^{{{\log }_{e}}x}}}dx\] \[\Rightarrow \] \[{{I}_{1}}=x{{.10}^{{{\log }_{e}}x}}-\int{x{{.10}^{{{\log }_{e}}x}}.\frac{{{\log }_{e}}10}{x}}dx\] \[\Rightarrow \] \[{{I}_{1}}=x{{10}^{{{\log }_{e}}x}}-\int{{{10}^{{{\log }_{e}}x}}{{\log }_{e}}10\,}dx\] \[\Rightarrow \] \[(1+{{\log }_{e}}10){{I}_{1}}=x{{10}^{{{\log }_{e}}x}}\] \[\Rightarrow \] \[{{I}_{1}}=\frac{x{{.10}^{{{\log }_{e}}x}}}{1+{{\log }_{e}}10}\] \[\therefore \] \[I=\left[ \frac{x{{.10}^{{{\log }_{e}}x}}}{1+{{\log }_{e}}10} \right]_{1}^{e}\] \[=\left[ \frac{10e-1}{1+{{\log }_{e}}10} \right]\] \[=\frac{10e-1}{{{\log }_{e}}10e}\]
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