Fundamental Theorem of Calculus

Mean value theorem for integrals

Let $f (x)$ be continuous and on $[𝒶, 𝒷]$ . Then there is at least one point $ω \in [𝒶, 𝒷]$ such that

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

Find $f_{ave}$ of $f (x) = x + 1$ and $ω \in [1, 6]$ such that $f (ω) = f_{ave}$ .

Find $\int_{-1}^{2} x^{2} = 3$ , find $ω \in [-1, 2]$ such that $f (ω) = f_{ave}$ .

Let $f (t) = - \frac{t + 1}{2} + 4 = - \frac{t}{2} + \frac{7}{2}$ . Find the area $A (x)$ under the graph of $f (t)$ from $t = 1$ to $t = x$

A (x) = \int_{1}^{x} f (t) ⅆ t

where $x \in [1, 6]$ .

Let $f (x)$ be continuous on $[𝒶, 𝒷]$ then

F (x) = \int_{𝒶}^{x} f (t) ⅆ t x \in [𝒶, 𝒷]

is continuous on $[𝒶, 𝒷]$ and differentiable on $(𝒶, 𝒷)$ with derivative

F^{'} (x) = f (x)

\begin{matrix} F_{1} (x) & = & \int_{1}^{x} \frac{1}{t^{3} + 1} ⅆ t \\ F_{2} (x) & = & \int_{-1}^{ℯ^{x}} t ⅆ t & F_{3} (x) & = & \int_{-1}^{x} ℯ^{t} ⅆ t \\ F_{4} (x) & = & \int_{\sqrt{x}}^{5} sin t ⅆ t \\ F_{5} (x) & = & \int_{5 x}^{x} t^{3} ⅆ t \end{matrix}

If $f (x)$ is continuous on $[𝒶, 𝒷]$ and $F (x)$ is any antiderivative of $f (x)$ i.e., $F^{'} (x) = f (x)$ then

\int_{𝒶}^{𝒷} f (t) ⅆ t = F (𝒷) - F (𝒶)

\begin{matrix} \int_{-2}^{2} x^{2} ⅆ x & = \\ \int_{1}^{9} \frac{-1}{x^{2}} ⅆ x & = \\ \int_{-1}^{4} \frac{-1}{x^{2}} ⅆ x & = \end{matrix}