Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
Consider a real valued continuous function $f$ such that $f(x) = \sin x + \int_{-\pi/2}^{\pi/2} (\sin x + tf(t)) dt$, then minimum value of $f(x)$ is____.
Step-by-Step Solution
Key Concept: Functional equations involving integrals are solved by treating the integral as a constant that can be determined through self-consistency.
Given $f(x) = \sin x + \int_{-\pi/2}^{\pi/2} (\sin x + tf(t)) dt$, let $A = \int_{-\pi/2}^{\pi/2} (\pi + 1)\sin t + A) dt = 2(\pi + 1) \int_0^{\pi/2} t \sin t dt$. Computing $A = 2(\pi + 1)$ by integration by parts, we find $f(x) = (\pi + 1)\sin x + 2(\pi + 1)$. Thus $f_{\max} = 3(\pi + 1)$ and $f_{\min} = -(\pi + 1)$.
Correct Answer: 4.14