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Question #41514

Difficulty: 8/10 | 🟢 Free

The Question

Let $I_n(\alpha) = \int_0^\alpha x^{2n+1} \ln(\alpha^2 - x^2) dx$ for $\alpha > 0$ and $n \in \mathbb{N}$ . Which of the following represents the value of $I_4'(e)$ ?

Options

A.
$-\frac{e^9}{12}$
B.
$\frac{e^9}{12}$
C.
$-\frac{e^9}{60}$
D.
$\frac{137e^9}{60}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Reduction Formulae|Leibniz Rule for Differentiation under Integral|Logarithmic Differentiation Domain

Why This Question Is Tricky

Common Mistakes

This question tests the 'Logarithmic Differentiation Domain' trap within the context of the Leibniz rule.

Students commonly apply the Leibniz rule blindly by differentiating the integrand with respect to the parameter $\alpha$ .

However, the integrand $f(x,\alpha) = x^{2n+1} \ln(\alpha^2 - x^2)$ is singular at the upper limit $x=\alpha$ , causing the boundary term $\ln(0)$ to be undefined and the resulting improper integral to diverge.

This creates an indeterminate form $-\infty + \infty$ .

To correctly solve the problem, one must first evaluate the integral using a reduction formula and integration by parts with a carefully chosen constant of integration to eliminate the singular boundary term, and only then differentiate the resulting closed-form expression.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Reduction Formulae
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free