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)
Explanation
Integrate by parts with \(u = \ln t\) and \(dv = t \, dt\):
\[\int t \ln t \, dt = \frac{t^{2}}{2}\ln t - \frac{t^{2}}{4}.\]
So \(\int_{1}^{x} t \ln t \, dt = \frac{x^{2}}{2}\ln x - \frac{x^{2}}{4} + \frac{1}{4}\). Setting this equal to \(\frac{1}{4}\) gives \(\frac{x^{2}}{2}\ln x = \frac{x^{2}}{4}\). Since \(x \neq 0\), \(\ln x = \frac{1}{2}\), so \(x = \sqrt{e}\).
The other options fail this last check: \(e\), \(e^{2}\) and \(e - 1\) give \(\ln x = 1\), \(2\) and about \(0.54\), not \(\frac{1}{2}\).
Answer: A