A) \[\frac{{{x}^{2}}}{2}\ell n\,\,x+c\]
B) \[-\frac{{{x}^{2}}}{4}\ell n\,\,x+\frac{{{x}^{2}}}{2}+c\]
C) \[\frac{{{x}^{2}}}{2}\ell n\,\,x+\frac{{{x}^{2}}}{4}+c\]
D) \[\frac{{{x}^{2}}}{2}\ell n\,\,x-\frac{{{x}^{2}}}{4}+c\]
Correct Answer: D
Solution :
[D]| \[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{x}^{1/n}}-{{x}^{1/(n+1)}}}{\frac{1}{{{n}^{2}}}}\] |
| \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{x}^{1/n}}\left( 1-{{x}^{1/n(n+1)}} \right]}{\left[ -\frac{1}{n(n+1)} \right]}.\,\,\,\,\,\left[ -\frac{{{n}^{2}}}{n(n+1)} \right]\] |
| Using \[L-H\] rule |
| \[=(-\,{{\log }_{e}}(x))\,\,(-1)=+{{\log }_{e}}x\] |
| \[\int{x.f(x)\,\,dx=\int{x.(lo{{g}_{e}}x)\,\,dx}}\] |
| \[={{\log }_{e}}x.\frac{{{x}^{2}}}{2}-\int{\frac{1}{x}.\frac{{{x}^{2}}}{2}dx={{\log }_{e}}x.\frac{{{x}^{2}}}{2}-\frac{{{x}^{2}}}{4}+C}\] |
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