Question

If ƒ(x) = $\lim_{n \rightarrow \infty}$n² (x^1/n – x^1/(n+1)), x > 0 then $\int_{}^{}x$ƒ(x) dx is equal to –

• A. x²/2
• B. 0
• C. x² log x – $\frac{1}{2}x^{2}$+ c
• D. None

Answer 1

Answer: (D)

None

Answer 2

ƒ(x) =$\lim_{n \rightarrow \infty}$n² (x^1/(n+1))

=$\frac{x^{\frac{1}{(n + 1)}}\left( x^{\frac{1}{n(n + 1)}} - 1 \right)}{\frac{1}{n(n + 1)} \times \frac{n(n + 1)}{n^{2}}}$ = log x.

Hence ƒ(x) dx = $\int_{}^{}x$log x dx = $\frac{x^{2}}{2}$log

x – $\frac{1}{4}$x² + c.