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Question #41576

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \int_{0}^{x} \frac{4}{5}\sqrt{25 - t^2} dt$ for $x \in [0, 5]$ . Let $g(x)$ be the inverse function of $f(x)$ . Evaluate the integral: $I = \int_{0}^{5\pi} \frac{x \cdot g(x)}{\sqrt{25 - (g(x))^2}} dx$

Options

A.
$\frac{75\pi}{2}$
B.
$\frac{125\pi}{4}$
C.
$50\pi$
D.
$\frac{25\pi}{2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Standard Ellipse/Circle Area|Inverse Function Integration|Inverse Function Derivative|Parametric Derivative Endpoint Trap

Why This Question Is Tricky

Common Mistakes

This question tests the integration of inverse functions where the derivative of the inverse function becomes undefined at an endpoint (the 'Parametric Derivative Endpoint Trap').

The integral is improper at $x = 5\pi$ because $g(5\pi) = 5$ , causing the denominator to vanish.

Students often fail to recognize that substituting $x = f(u)$ perfectly regularizes the integral by canceling the vanishing denominator.

Attempting to evaluate the integral using integration by parts without this transformation leads to undefined boundary terms or incorrect limit evaluations.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Standard Ellipse/Circle Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free