Discussion
The value of \(x\) such that \(x \gt 1\), satisfying the equation \(\int_{1}^{x} t\ln t\,dt = \frac{1}{4}\) is
Answer: A
The value is \(x = \sqrt{e}\).
Integrate by parts, with u = ln t and dv = t dt:
\[ \begin{aligned} \int t\ln t\,dt &= \frac{t^{2}}{2}\ln t - \frac{t^{2}}{4} \\ \int_{1}^{x} t\ln t\,dt &= \frac{x^{2}}{2}\ln x - \frac{x^{2}}{4} + \frac{1}{4} \end{aligned} \]
Setting this equal to \(\frac{1}{4}\) leaves \(\frac{x^{2}}{2}\ln x = \frac{x^{2}}{4}\), so \(\ln x = \frac{1}{2}\) and \(x = e^{1/2} = \sqrt{e}\).
\(x = e\) gives \(\frac{e^{2}}{2} - \frac{e^{2}}{4} + \frac{1}{4} = \frac{e^{2}+1}{4}\), far more than \(\frac{1}{4}\); \(e^{2}\) is larger still, and \(e - 1\) does not make \(\ln x = \frac{1}{2}\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development