Let \(F(x) = \int_{1}^{x^{2}}\displaystyle\log_e t\,dt\) for \(x \gt 0\). Then \(F'(e)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4e
By the first fundamental theorem of calculus together with the chain rule, if \(F(x) = \int_{a}^{u(x)} g(t)\,dt\), then \(F'(x) = g(u(x))\,u'(x)\).
Here \(g(t) = \log_e t\) and \(u(x) = x^{2}\), so
\(F'(x) = \log_e(x^{2})\cdot 2x\)
\( = 4x\log_e x\)
At \(x = e\):
\(F'(e) = 4e\log_e e = 4e\)
Why the other options are incorrect
Option (2) 2e — Incorrect. It puts t = x instead of \(t = x^{2}\) in \(\log_e t\), giving \(2e\log_e e\).
Option (3) 2 — Incorrect. This is only \(\log_e(e^{2})\); the factor \(2x = 2e\) from the chain rule is left out.
Option (4) \(e^{2}\) — Incorrect. This is the upper limit \(x^{2}\) at \(x = e\), not the derivative.