Integration bound
Dependencies: None
Let $f: \mathbb{R} \mapsto \mathbb{R}$ be a non-increasing function, i.e. $x < y \implies f(x) \ge f(y)$. Then \[ \int\limits_a^{b+1} f(x)dx \,\le\, f(b) + \int\limits_a^b f(x)dx \,\le\, \sum_{i=a}^b f(i) \,\le\, f(a) + \int\limits_a^b f(x)dx \,\le\, \int\limits_{a-1}^b f(x)dx \]
Proof
\begin{align} & \int_a^b f(x)dx \\ &= \sum_{i=a}^{b-1} \int_i^{i+1} f(x)dx \\ &= \sum_{i=a}^{b-1} \int_i^{i+1} [f(i+1), f(i)] dx \\ &= \sum_{i=a}^{b-1} [f(i+1), f(i)] \\ &= \left[ \sum_{i=a}^{b-1} f(i+1), \sum_{i=a}^{b-1} f(i) \right] \\ &= \left[ \sum_{i=a+1}^b f(i), \sum_{i=a}^{b-1} f(i) \right] \\ &= \left[ \left( \sum_{i=a}^b f(i) \right) - f(a), \left( \sum_{i=a}^b f(i) \right) - f(b) \right] \\ &= \left( \sum_{i=a}^b f(i) \right) - \left[ f(b), f(a) \right] \end{align} \[ \int_a^{b+1} f(x)dx - \int_a^b f(x)dx = \int_b^{b+1} f(x)dx \le \int_b^{b+1} f(b)dx = f(b) \] \[ \int_{a-1}^b f(x)dx - \int_a^b f(x)dx = \int_{a-1}^a f(x)dx \ge \int_{a-1}^a f(a)dx = f(a) \]
Dependency for:
Info:
- Depth: 0
- Number of transitive dependencies: 0