Definite Integration
Fractional part of complex integral sum
MJAT_TS8_P1
Grade 12
Question:
Let $f(x)=\frac{1}{2}\!\left(\tan^{-1}\!\left(x-\frac{1}{2}\right)+\{3-x\}+\{5+x\}\right)$ and $I=\displaystyle\sum_{r=0}^{2022}\int_1^{2025}f\!\left(x+\frac{r}{2025}\right)dx=\frac{b}{a}\cdot 2025$ where $\gcd(a,b)=1$. If $\left\{\frac{b}{a}\right\}=\frac{2023}{x}$, find $x$ (where $\{\cdot\}$ denotes fractional part):
Step-by-Step Solution
Key Concept: Using the periodicity and symmetry: $f(x)+f(1-x)=\{x\}+\{-x\}$ (from the arctan part cancelling via $f(x)+f(1-x)$). The sum telescopes over the 2023 terms. $I=\int_1^{2024}f(x)dx$ effectively.
$x=\mathbf{4}$.
Correct Answer: 4