The definite integral (5.2 and 5.3)

• Graphically: the area between the curve and the $x$-axis, from $a$ to $b$
• The limit as the number of sub-intervals, $n$, increases $\to \infty$ of our sum of rectangles (either a left sum or a right sum).

A recipe for left- and right-sums

Left sum

1. Divide the $x$ interval from $x=a$ to $x=b$ up into $n$ sub intervals. The width of each sub-interval (of each rectangle) will be $$\Delta = \frac{b-a}{n}.$$
2. Draw a rectangle for each interval with the width of $\Delta$, and a height equal to $f(x_\text{left})$ the value of the function at the left-most $x$-value in each interval:
   
   
3. For a left sum, the sum of the rectangle areas comes to: $$\begineq \text{Left sum}=&f(a)\Delta+f(a+\Delta)\Delta+f(a+2\Delta)\Delta+....+ f(a+(n-1)\Delta)\Delta\\ =&\Delta\left[f(a)+f(a+\Delta)+f(a+2\Delta)+....+ f(a+(n-1)\Delta)\right]\\ =&\Delta\sum_{i=0}^{n-1} f(a+i\Delta) \endeq $$

   Right sum

   1. Divide the $x$ interval from $x=a$ to $x=b$ up into $n$ sub intervals. The width of each sub-interval (of each rectangle) will be $$\Delta = \frac{b-a}{n}.$$
   2. Draw a rectangle for each interval with the width of $\Delta$, and a height equal to $f(x_\text{right})$ the value of the function at the right-most $x$-value in each interval:
      
      
   3. For a right sum, the sum of the rectangle areas comes to: $$\begineq \text{Right sum}=& f(a+\Delta)\Delta+f(a+2\Delta)\Delta+....+ f(a+(n-1)\Delta)\Delta\\ =&\Delta\left[f(a+\Delta)+f(a+2\Delta)+....+ f(a+(n-1)\Delta)\right]\\ =&\Delta\sum_{i=1}^{n} f(a+i\Delta) \endeq $$

      Homework

      On problems like 5.2 #18, You can use the CoCalc notebook you developed in lab to find the values of "the definite integral" (as the limit with several thousand rectangles) when they don't tell you a particular number of rectangles to use.

      On problem 5.3 #3...
      With the definite integral, areas below the x-axis count negative and areas above the x-axis count positive. But this problem asks for the "total area" between the function and the x-axis. The answer they're looking for is not the definite integral over that interval, but instead the total, absolute value of areas whether below or above the axis. Figure out where the function crosses the x-axis. Then, use CoCalc to figure out the area that is above the x-axis, and separately the area that is below the x-axis (this is a negative area). When I added the absolute values of those two together Wiley marked my answer correct.

      Which of the following is the best estimate for $$\int_0^9 f(t)\,dt?$$

      a) 13   b) 30   c) 3   d) 18   e) 9
      

      The integral represents this area...
      
      
      which is more than the area of this triangle = $\frac 12*9*3=13.5$
      
      
      and which is less than the area of this rectangle = $9*3=27$
      
      
      The only answer that satisfies $13.5 \le A \le 27$ is choice d) 18

      For this graph:
      
      

      1. Give an interval on which the left-hand sum approximation of the area under the curve on that interval is an underestimate.-3 < $x$ < -1
      2. Give an interval on which the left-hand sum approximation of the area under the curve on that interval is an overestimate. 0 < $x$ < 3

      By the way, how do we interpret the "area under the curve" if $f(x)$ is negative? If we think of velocity: positive velocity means going away from our starting point, and negative velocity means coming back towards it. So, count the area between the $x$ axis and $f(x)$ as negative when $f(x)$ is negative.

      For which of the graphs below is the definite integral $$\int_0^{10} f(x)\,dx$$ closest to zero?
      

      
      For each graph, add the pink areas (areas above the $x$-axis count as positive) and subtract the blue areas (areas below the $x$-axis count as negative). In graph b it looks the pink just balances blue, summing to zero.

      A bicyclist starts from home and rides back and forth along a straight east/west highway. Her velocity is given in the graph. Positive velocities indicate travel toward the east, negative toward the west.
      

      1. On what time intervals is she stopped? Stopped for $3 < t < 5$ and $9 < t < 10$.
      2. How far from home is she the first time she stops, and in what direction?

         The first time she stopped was at $t=9$. Up to that point she had travelled an amount corresponding to the area in pink:
         
         A problem is that the area has units of $x$ units times $y$ units=minutes * (ft/sec), and this is awkward. One solution is to convert the speed to feet/min, like this: 30 ft/sec=30ft/sec*60 sec/min = 1800 ft/min. (or Google: '30 feet/sec in feet/min')
         
         So, now we can add up the pink area above, as the sum of a triangle, a rectangle, and a triangle shown. That pink area = 1 min * 1800 ft/min /2 +1 *1800 + 1*1800/2= = 900 feet + 1800 + 900= 3600 feet (about 2/3 of a mile).

      3. At what time does she bike past her house?

         By the time she gets to $t=$7.5 seconds the negative 'blue' area = 1*1800/2+1.5*1800=900+2700= 3600 feet. And since it's below the $x-$axis, this cancels the pink area exactly.