Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
For differentiable function $f(x)$, if $\int_0^a f'(x)\left([x]-x+\frac{1}{2}\right) dx = A_1 \int_0^a f(x) dx + A_2 f(0) + A_3 f(a) + A_4 \sum_{r=0}^{a} f(r)$, (where $[.]$ Denotes the G.I.F and $A_1, A_2, A_3, A_4$ are constant $n \in \mathbb{N}$) then $A_1 + A_2 + A_3 + A_4$ is equal to ____.
Step-by-Step Solution
Key Concept: Partitioning the domain at integer boundaries and using telescoping sums simplifies integrals involving the floor function.
The sum $I = \int_0^n f'(x)[x]dx$ is evaluated by partitioning the integral at integer points and applying summation formulas. Using the property that $[x]$ is constant on intervals $[r, r+1)$, we expand the integral as $\sum_{r=1}^{n}(r-1)\int_r^{r+1} f'(x)dx$. After applying integration by parts and telescoping, we obtain $I = -\sum_{r=1}^n f(r) + \frac{1}{2}f(n) + \frac{1}{2}f(0) + \int_0^n f(x)dx$.
Correct Answer: 1