Definite Integration
King's Property with Functional Equation
nta_pyq_2023_apr
Grade 12
Question:
Let $f(x)$ be a function satisfying $f(x)+f(\pi-x)=\pi^2$, $\forall x\in\mathbb{R}$. Then $\displaystyle\int_0^\pi f(x)\sin x\,dx$ is equal to
\dfrac{\pi^2}{4}
2\pi^2
\pi^2
\dfrac{\pi^2}{2}
Step-by-Step Solution
Key Concept: Apply King's property: $I=\int_0^\pi f(\pi-x)\sin x\,dx$. Add both integrals using $f(x)+f(\pi-x)=\pi^2$.
$2I=\int_0^\pi[f(x)+f(\pi-x)]\sin x\,dx=\pi^2\int_0^\pi\sin x\,dx=2\pi^2$. $I=\pi^2$.
Correct Answer: 3