The value of \(x\) such that \(𝑥 > 1\), satisfying the equation \(\int_1^x t…
2025
The value of \(x\) such that \(𝑥 > 1\), satisfying the equation \(\int_1^x t \ln t \, dt = \frac 1 4\) is
- A.
\(\sqrt{e}\) - B.
\(e\) - C.
\(e^2\) - D.
\(𝑒 − 1\)
Attempted by 41 students.
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Correct answer: A
Solution:
Compute the integral by parts.
Let u = ln t and dv = t dt. Then du = dt/t and v = t^2/2.
An antiderivative is (t^2/2) ln t - t^2/4. Evaluating from 1 to x gives (x^2/2) ln x - x^2/4 + 1/4.
Set this equal to 1/4 and simplify: (x^2/2) ln x - x^2/4 = 0, which is (x^2/4)(2 ln x - 1) = 0.
For x > 1 we have x^2/4 ≠ 0, so 2 ln x - 1 = 0. Thus ln x = 1/2 and x = e^{1/2} = sqrt(e).
Therefore the required value of x (with x > 1) is sqrt(e).
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