Riemann sums and the definite integral

Goal: compute total amount of change over a period of time.

Walking speed: $\displaystyle \diff{x}{t} = 1$

Walking speed: $\displaystyle \diff{x}{t} = 1 + 3t$

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Distance = rate $\times \Delta t$

Let $f(t)=1+3t$
         $t_i = i\Delta t$,    for $ i=1, 2, \ldots, n$.


Rate=`{}`
Distance: `{}`

Total distance: `{}`

Riemann sums
$$I_l = \sum_{i=1}^n f(t_{i-1})\Delta t$$ Left Riemann sum

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$$I_r = \sum_{i=1}^n f(t_{i})\Delta t$$ Right Riemann sum

Define definite integral (Riemann integral) as limit as $n \to \infty$.   ($\Delta t = 2/n$)

\begin{align*} \text{Amount walked} &= \int_0^2 f(t) dt\\ &= \lim_{n \to \infty} \sum_{i=1}^n f(t_{i-1})\Delta t\\ &= \lim_{n \to \infty} \sum_{i=1}^n f(t_{i})\Delta t \end{align*}

Definite integral

The definite integral of the function $f(t)$ over the interval $t \in [a,b]$:

\begin{align*} \int_a^b f(t) dt &= \lim_{n \to \infty} \sum_{i=1}^n f(t_{i-1})\Delta t\\ &= \lim_{n \to \infty} \sum_{i=1}^n f(t_{i})\Delta t \end{align*}
where $\displaystyle\Delta t = \frac{b-a}{n}$ and $t_i = a + i \Delta t$.

Example: estimate $\displaystyle \int_{-4}^{-1} x^2 dx$ with left and right Riemann sums of 6 intervals.

$a=-4$, $b=-1$, $\Delta x =(b-a)/6 = (-1-(-4))/6 = 3/6 = 1/2$

$x_0 = -4$, $x_1=-3.5$, $x_2=-3$, $x_3=-2.5$, $x_4=-2$, $x_5=-1.5$, $x_6=-1$

Intervals: $[-4,-3.5]$, $[-3.5, -3]$, $[-3, -2.5]$, $[-2.5, -2]$, $[-2, -1.5]$, $[-1.5,-1]$

$\displaystyle I_l = \sum_{i=1}^6 (x_{i-1})^2 \Delta x$ $= x_0^2\Delta x + x_1^2\Delta x + x_2^2\Delta x + x_3^2\Delta x + x_4^2\Delta x + x_5^2\Delta x$
  $=(-4)^2 0.5+ (-3.5)^2 0.5 + (-3)^2 0.5 + (-2.5)^2 0.5 + (-2)^2 0.5 + (-1.5)^2 0.5$
  $=24.875$

$\displaystyle I_r = \sum_{i=1}^6 (x_{i})^2 \Delta x = x_1^2\Delta x + x_2^2\Delta x + x_3^2\Delta x + x_4^2\Delta x + x_5^2\Delta x + x_6^2\Delta x$
  $=(-3.5)^2 0.5 + (-3)^2 0.5 + (-2.5)^2 0.5 + (-2)^2 0.5 + (-1.5)^2 0.5 + (-1)^2 0.5$
  $=17.375$