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I have been trying to figure out how we would find the limit of the the definite integral exactly, and I was thinking back to how we found limits before (or derivatives of a function. Would it be possible to "solve" the equation with a theorem or possible to add or subtract the anti-derivatives at the two point since that would be showing the "original" function of that derivative which is the area/distance.

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I was wondering if or when we might get to explore something more like this so that we could get a different look at derivatives.

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The graph of problem B on the homework seems to have symmetrical points across x=6. Though this might not be the case, I cannot distinguish the difference. Thus, can I simply use the heights for the rectangles on one side for the other? Provided I account for the middle and the edge.

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This may be a little far fetched but I was wondering, if similar to anti-derivitives there was a way to find the original definite integral from the answer of the Riemann Sum.
Do definite integrals in any way help us find anti-derivitves-- are they at all related?

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If you did a right Riemann sum and a left Riemann sum and averaged those together would that be fairly accurate? I explored on geogebra and found in the equation I used that the midpoint Riemann sum was closer to the actual integral by 0.07 , but would that always be the case? Is this a strategy that people use because it seems more accurate than just using a left or right Riemann sum, but I couldn't find anything on the internet?

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I've noticed that all of the examples take place in the first quadrant and sometimes the fourth quadrant. Is it possible to find the definite integral when the interval is between negative x-coordinates? If it is possible, is the area in the second quadrant negative and the area in the third positive? I can see how with the x-axis being time it might not make sense, but I don't really see a reason why the area in the second and third quadrants couldn't be found in the same way as the other two if the function is defined.

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The definite integral is found by the area between the x-axis and the graph of the function. We know that this has a real life context (distance, volume, etc.) but what about the space above the graph? Can that tell us anything? Can we even find what that area would be since the graph is on a plane, that doesn't end? I did a little research and all I could find is about when the graph is below the x-axis. I did find the linked website to be helpful and interesting.

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Content

Areas above and below the axis

So far we have only considered functions \(f(x)\) whose graphs are above the \(x\)-axis, i.e., \(f(x) > 0\). But, of course, a graph can go below the \(x\)-axis. We now consider this situation.

When we computed the right-endpoint estimate, we approximated the area under the graph \(y=f(x)\) by rectangles of width \(\Delta x = \frac{1}{n} (b-a)\) and height \(f(x_j)\). The important point now is that, if \(f(x_j)\) is negative, then \(f(x_j) \; \Delta x\) is negative and we obtain the negative of the area of the rectangle. So when the graph of \(y=f(x)\) is below the \(x\)-axis, our method approximates the negative of the area between the graph and the \(x\)-axis.

In regions where the graph of \(y=f(x)\) is above the \(x\)-axis (i.e., \(f(x)>0\)), a definite integral calculates the area between the graph and the \(x\)-axis. In regions where the graph is below the \(x\)-axis (i.e., \(f(x) < 0\)), the integral calculates the negative of the area between the graph and the \(x\)-axis. A general function \(f(x)\) is sometimes positive and sometimes negative, so the integral calculates the signed area, that is, the total area above the \(x\)-axis minus the total area below the \(x\)-axis.

Detailed description of diagram
The integral counts areas above/below the x-axis as positive/negative.

When we have a velocity-time graph \(y=v(t)\), the signed area has a natural interpretation. The regions where the graph is below the \(t\)-axis occur when \(v(t) < 0\), i.e., the velocity is negative and you are going backwards. When you are going backwards, your change in position is negative. Your total displacement is given by your total forwards movements minus your total backwards movements; these correspond to the regions above and below the \(t\)-axis. Thus signed area corresponds to displacement. The total area enclosed between the graph and the axis, on the other hand, is the total distance travelled, forwards and backwards.

Detailed description of diagram
In general, to find the total area enclosed between a graph and the axis, we find where the graph crosses the axis and compute the areas separately. See the following example.

Example

1. Find \(\displaystyle\int_0^2 (-x^2 - x + 2) \; dx\).

2. Find the area of the shaded region.

3. If your velocity at time \(t\) is given by \(v(t) = -t^2 - t + 2\), then what is your displacement and your distance travelled between time \(t=0\) and \(t=2\)?

Solution
Screencast of exercise 12 Next page - Content - Properties of the definite integral

1. \begin{align*} \int_0^2 (-x^2 - x + 2) \; dx &= \Bigg[ {-} \frac{1}{3} x^3 - \frac{1}{2} x^2 + 2x \Bigg]_0^2\\ &= \Big( {-} \frac{8}{3} - 2 + 4 \Big) - \big( 0 \big) = - \frac{2}{3}. \end{align*}

2. The area is partly above and partly below the \(x\)-axis. Factorising \[ -x^2 - x + 2 = -(x-1)(x+2), \]
we find the \(x\)-intercept is at \(x=1\). Therefore the total area shown is
\begin{gather*} \int_0^1 (-x^2 - x + 2) \; dx - \int_1^2 (-x^2 - x + 2) \; dx \\ \qquad\qquad\begin{aligned}[t] &= \Bigg[ {-} \frac{1}{3} x^3 - \frac{1}{2} x^2 + 2x \Bigg]_0^1 - \Bigg[ {-} \frac{1}{3} x^3 - \frac{1}{2} x^2 + 2x \Bigg]_1^2 \\ &= \Bigg( \Big( {-} \frac{1}{3} - \frac{1}{2} + 2 \Big) - \big( 0 \big) \Bigg) - \Bigg( \Big( {-} \frac{8}{3} - 2 + 4 \Big) - \Big( {-} \frac{1}{3} - \frac{1}{2} + 2 \Big) \Bigg) \\ &= \Big( \frac{7}{6} - 0 \Big) - \Big( {-} \frac{2}{3} - \frac{7}{6} \Big) \\ &= \frac{7}{6} - \Big( {-} \frac{11}{6} \Big) = 3. \end{aligned} \end{gather*}

3. Your displacement is \(\int_0^2 v(t) \; dt\), which was computed in part 1, and is \(-\frac{2}{3}\) (i.e., you end up \(\frac{2}{3}\) backwards of where you started). Your total distance travelled is given by the area calculated in part 2, which is 3.

Beware! Computing integrals often involves many fractions, subtractions and negative signs, as in the above example! It is good practice to carefully bracket all terms, as shown.

Exercise 12

1. Compute \(\displaystyle\int_{-2}^2 (x^3 - x) \; dx\).

2. What is the total area enclosed between the graph of \(y=x^3 - x\) and the \(x\)-axis, from \(x=-2\) to \(x=2\)?

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Content - Areas above and below the axis

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So far we have only considered functions \(f(x)\) whose graphs are above the \(x\)-axis, i.e., \(f(x) > 0\). But, of course, a graph can go below the \(x\)-axis. We now consider this situation. When we computed the right-endpoint estimate, we approximated the area under the graph \(y=f(x)\) by rectangles of width \(\Delta x = \frac{1}{n} (b-a)\) and height \(f(x_j)\).

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I did this by doing a right hand Riemann Sum to start but the wording in the 3rd part, implies that we weren't supposed to do it until that part so I was wondering what we were supposed to have done in the first part or what other people did?

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I would guess just from looking that this would be more accurate but according to the website I found this is not as accurate as the midpoint rule.

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I don't know how to make more partitions!! But The mid point rule is cool.

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I don't know how to make more partitions!! But The mid point rule is cool.

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So the last extra extra thing on the assignment is beyond me but I did find something cool called the midpoint rule. Basically instead of taking the right or left hand sum you sum the area of the rectangle formed by the midpoint of your boundaries. which if the concavity of the curve doesn't change is the most accurate way to estimate the area under it. Like this

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PDF or other files can get posted. Make sure everything is appropriate academic work

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You can post pics of work to help others or to ask questions of the group

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With the three dot menu selector, you can choose Draw to write things on a touchscreen

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Here is some calculator help aimed at the extension problem on the exploration but maybe useful elsewhere