Riemann integral

Let $f (x)$ be defined on $[𝒶, 𝒷]$ . We say $𝒥$ is the definite (Riemann) integral of $f (x)$ over $[𝒶, 𝒷]$ if

\begin{matrix} \forall ε \in ℝ^{+}, \exists δ (ε) \in ℝ^{+} : \\ \forall 𝒫, | 𝒫 | < δ (ε), 𝒸_{i} \in [χ_{i - 1}, χ_{i}] \Rightarrow | 𝒥 - \sum_{κ = 1}^{n} f (𝒸_{i}) Δ χ_{i} | < ε \end{matrix}

\begin{matrix} \int_{𝒶}^{𝒷} f (x) ⅆ x & = & lim_{n \to \infty} \sum_{κ = 1}^{n} f (𝒸_{i}) Δ χ \\ = & lim_{n \to \infty} \sum_{κ = 1}^{n} f (𝒸_{i}) \frac{𝒷 - 𝒶}{n} \\ = & lim_{n \to \infty} \sum_{κ = 1}^{n} f (𝒶 + κ \frac{𝒷 - 𝒶}{n}) \frac{𝒷 - 𝒶}{n} \\ = & \dots \end{matrix}

If $f (x)$ is continuous on $[𝒶, 𝒷]$ , or has at most finitely many jump discontinuities then $f (x)$ is integrable over $[𝒶, 𝒷]$ .

Area

If $f (x)$ is non-negative and integrable over $[𝒶, 𝒷]$ , then the area under the curve $y = f (x)$ over the interval $[𝒶, 𝒷]$ is

𝒜 = \int_{𝒶}^{𝒷} f (x) ⅆ x

Evaluate $\int_{0}^{1} f (x) ⅆ x$ for

\begin{matrix} f (x) & = & x^{2} \\ f (x) & = & {\begin{matrix} 1, & x \in ℚ \\ 0, & x \in ℝ ∖ ℚ \end{matrix} \end{matrix}

Find the area under the curve

f (x) = \sqrt{9 - {(x - 4)}^{2}}

f (x) = x - 2

over $[-3, 3]$ , $[-1, 5]$ and $[0, 6]$ .

Suppose

\int_{0}^{8} f (x) ⅆ x = 10 \int_{0}^{5} f (x) ⅆ x = 3

Evaluate

\int_{0}^{8} f (x) ⅆ x

Estimate

\int_{0}^{1} ℯ^{- x^{2}} ⅆ x

Find the average value of $f (x) = x + 1$ on $[1, 6]$

Find the average value of

f (x) = \sqrt{9 - {(x - 3)}^{2}}

\begin{matrix} 1. & \int_{𝒶}^{𝒷} 𝒸 ⅆ x = 𝒸 (𝒷 - 𝒶) \\ 2. & \int_{𝒶}^{𝒷} (α f (x) + β g (x)) ⅆ x = α \int_{𝒶}^{𝒷} f (x) ⅆ x + β \int_{𝒶}^{𝒷} g (x) ⅆ x \end{matrix}

\begin{matrix} 1. & \int_{𝒶}^{𝒶} f (x) ⅆ x = 0 \\ 2. & \int_{𝒶}^{𝒷} f (x) ⅆ x = - \int_{𝒷}^{𝒶} f (x) ⅆ x \\ 3. & \int_{𝒶}^{𝒸} f (x) ⅆ x = \int_{𝒶}^{𝒷} f (x) ⅆ x + \int_{𝒷}^{𝒸} f (x) ⅆ x \end{matrix}

If $g (x) \leq f (x)$ for every $x$ in the interval $[𝒶, 𝒷]$ then

\int_{𝒶}^{𝒷} g (x) ⅆ x \leq \int_{𝒶}^{𝒷} f (x) ⅆ x

If $𝓁 \leq f (x) \leq 𝓊$ for every $x$ in the interval $[𝒶, 𝒷]$ then

𝓁 (𝒷 - 𝒶) \leq \int_{𝒶}^{𝒷} f (x) ⅆ x \leq 𝓊 (𝒷 - 𝒶)

Let $f (x)$ be integrable on $[𝒶, 𝒷]$ . The average value of the function on the interval is

f_{ave} = \frac{1}{𝒷 - 𝒶} \int_{𝒶}^{𝒷} f (x) ⅆ x