### Integrals and the Fundamental Theorem of Calculus

Indefinite Integral

$\displaystyle \int f(t) dt$ $\displaystyle = F(t) + C$

$$\diff{F}{t} = f(t)$$

Definite Integral

$\displaystyle \int_a^b f(t) dt$   $=$
total change
from $t=a$ to $t=b$
with rate of change $f(t)$

An anti-derivative

A function (with arbitrary constant $C$)

The limit of Riemann sums

A number

A fundamental relationship between the indefinite integral and definite integral:
The Fundamental Theorem of Calculus

Indefinite Integral

$\int f(t) dt = F(t) + C$

$\diff{F}{t} = f(t)$

Definite Integral

$\int_a^b f(t) dt =$
total change
from $t=a$ to $t=b$
with rate of change $f(t)$

If $f(t)$ is walking speed at time $t$:

$\int f(t)dt =$ position at time $t$
($C$ determined by initial position)

$\int_a^b f(t) dt =$ distance walked from
time $a$ to time $b$.

The Fundamental Theorem of Calculus

The distance walked is the change in position.
The distance walked is final position minus initial position.

The definite integral is the change in the indefinite integral.

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= 5 \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 + C \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 + C = C = 5$

$X(t) = t + \tfrac{3}{2}t^2 +5$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + 5 - (5)$ $=13 - 5 = 8$

Walking example. Walking speed: $f(t)=1+3t$.

Recall Riemann sums

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Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + 5 - (5)$ $=13 - 5 = 8$

Moral of the
Fundamental Theorem of Calculus

If you can find the anti-derivative,
you can skip the Riemann sum.

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= \color{green}{\fbox{5}} \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 + C \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 + C = C = \color{green}{\fbox{5}}$

$X(t) = t + \tfrac{3}{2}t^2 +\color{green}{\fbox{5}}$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + \color{green}{\fbox{5}} - (\color{green}{\fbox{5}})$ $=(8 + \color{green}{\fbox{5}}) - \color{green}{\fbox{5}} = 8$

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= \color{green}{\fbox{7}} \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 + C \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 + C = C = \color{green}{\fbox{7}}$

$X(t) = t + \tfrac{3}{2}t^2 +\color{green}{\fbox{7}}$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + \color{green}{\fbox{7}} - (\color{green}{\fbox{7}})$ $=(8 + \color{green}{\fbox{7}}) - \color{green}{\fbox{7}} = 8$

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= \color{green}{\fbox{-11}} \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 + C \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 + C = C = \color{green}{\fbox{-11}}$

$X(t) = t + \tfrac{3}{2}t^2 +\color{green}{\fbox{-11}}$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + \color{green}{\fbox{-11}} - (\color{green}{\fbox{-11}})$ $=(8 + \color{green}{\fbox{-11}}) - (\color{green}{\fbox{-11}}) = 8$

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= \color{green}{\fbox{C}} \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 + C \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 + C = C = \color{green}{\fbox{C}}$

$X(t) = t + \tfrac{3}{2}t^2 +\color{green}{\fbox{C}}$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 + \color{green}{\fbox{C}} - (\color{green}{\fbox{C}})$ $=(8 + \color{green}{\fbox{C}}) - \color{green}{\fbox{C}} = 8$

For the definite integral,
you can ignore the constant $C$
in the anti-derivative.

Walking example. Walking speed: $f(t)=1+3t$.

Let $X(t)$ be position at time $t$.

\begin{align*} \diff{X}{t} &= 1 + 3t\\ X(0) &= 0 \end{align*}

\begin{align*} X(t) &= \int f(t)dt = \int (1+3t)dt\\ &= t + \tfrac{3}{2}t^2 \end{align*}

$X(0) = 0 + \tfrac{3}{2}(0)^2 = 0$

$X(t) = t + \tfrac{3}{2}t^2$

Distance walked from $t=0$ to $t=2$.

 $\displaystyle \int_0^2 f(t) dt$ $\displaystyle= \int_0^2 (1+3t) dt$ $= X(2) - X(0)$ $=2 +\tfrac{3}{2}(2)^2 - (0)$ $=8 - 0 = 8$

For the definite integral,
you can ignore the constant $C$
in the anti-derivative.

### The Fundamental Theorem of Calculus

Let $F(x)$ be any indefinite integral

$$F(x)=\int f(x)dx,$$

then

$$\int_a^b f(x) dx = F(b) - F(a).$$

$$\int_a^b f(x) dx = F(x)\big|_a^b$$

where $F(x)\big|_a^b =F(b) - F(a)$.

Example: Calculate $\displaystyle\int_{-1}^2 e^{3x}dx$

Step 1: Calculate the indefinite integral.

$\displaystyle \int e^{3x}dx = \frac{1}{3} \int 3 e^{3x}dx = \frac{1}{3} e^{3x}$ $+C$

Step 2: Use the Fundamental Theorem of Calculus

 $\displaystyle \int_{-1}^2 e^{3x} dx$ $\displaystyle= \frac{1}{3} e^{3x}\Big|_{-1}^2$ $\displaystyle= \frac{1}{3} e^{6} - \frac{1}{3} e^{-3}$ $= 134.47626 - 0.016596 \approx 134.46$