What do the bounds of a definite integral represent geometrically?
I understand they are the values of the integral evaluated at those points, but is there a visual geometric interpretation? Possibly one that relates to how subtracting the integral evaluated at those points and how that is equivalent to the area.
I've seen the images of shading underneath a curve to denote area, but what does the integral at a point, irrespective of anything else mean?
Put simply: I guess I'm confused as to how two values of a function one subtracted from another can produce the area underneath its derivative. I'm sure there are rigorous proofs, I would prefer a visual explanation as that's how I like to think.
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$\begingroup$You can think of the integral evaluated at a certain point as the area under the function's curve up until that point (where we call the area negative if the function is progressing below the $x$-axis and positive if it is progressing above the $x$-axis). So the integral evaluated at $b$ is the total area (under these sign conventions) under the curve up until $x=b$, and the integral evaluated at $a$ is the total area under the curve up until $x=a$. You can see intuitively from looking at the picture why subtracting the latter from the former would give us the shaded area. This is certainly not rigorous, but it may be the geometric intuition you are looking for.
$\endgroup$ 7 $\begingroup$Based on your explanation I assume you are talking about single variable integration. They are called "bounds" because they "bound" the portion of the curve you are talking about. You are evaluating the area of the region under the curve with $a\leq x \leq b$.
Another way of thinking about this is looking at the formal statement of the first part of the Fundamental Theorem of Calculus. It states that if
$F(x) = $ $$\int_{c}^{x} f(x) dx$$ then $F'(x) = f(x)$ (where $f$ is a continuous real valued function defined on $[a,b]$). $F(x)$ represents the area under the curve from the constant $c$ to $x$. You can think of $$\int_{a}^{b} f(x) dx$$ as $$\int_{c}^{b} f(x) dx - \int_{c}^{a} f(x) dx$$
(Note that this is $F(b) - F(a)$).
You can think of the above statement as the area under the curve from $c\leq x \leq b$ minus the area under the curve from $c\leq x \leq a$ the "overlapping area" is cancelled out in the subtraction and we are left with the area under the curve from $a\leq x\leq b$.
First of all, keep in mind that "area under the curve" is not a definition of an integral. It just happens to work out to the same value given some widely-applicable assumptions (and some added definitions such as what is "area under the curve" when the curve is below the $x$-axis).
Consider a reasonably "nice" curve given by a formula $y = f(x)$ that (for simplicity in measuring area) never dips below the $x$-axis. Among its "nice" properties, let's at least require $f$ to be continuous. For $t \geq 0$, define $F(t)$ as the area between the curve, the $x$-axis, the $y$-axis, and the line $x = t$.
Now increase $t$ by a very small amount, $\Delta t$. This adds a small strip to the right-hand side of the area we are measuring. The area of this strip is $F(t + \Delta t) - F(t)$, because the strip is just what is left over if you start with the entire region under the curve between the $y$-axis and $x = t + \Delta t$ (which has area $F(t + \Delta t)$ by definition) and then remove the region under the curve between the $y$-axis and $x = t$ (which has area $F(t)$).
The length of the left edge of the strip is $f(t)$, and since $f$ is continuous, as long as $\Delta t$ is small enough the top of the strip is all very close to the same distance from the $x$-axis, so the area of the strip is (approximately) $f(t) \times \Delta t$. The approximation is even better for smaller $\Delta t$; since $\lim_{x\to t} f(x) = f(t),$ we can show that the entire strip of area under $f$ between $t$ and $t + \Delta t$ is enclosed in a rectangle of width $\Delta t$ and height $f(t) + \varepsilon$, and encloses a rectangle of width $\Delta t$ and height $f(t) - \varepsilon$, and moreover by making $\Delta t$ small enough we can make $\varepsilon$ as close to zero as we like. We can therefore "squeeze" the area of this strip under $f$ between the areas of two rectangles, $$(f(t) - \varepsilon) \Delta t \leq (\text{area of strip}) \leq (f(t) + \varepsilon) \Delta t.$$
Recalling that the area of the strip is $F(t + \Delta t) - F(t)$, dividing by $\Delta t$, and then taking limits: $$\lim_{\Delta t \to 0} (f(t) - \varepsilon) \leq \lim_{\Delta t \to 0} \frac{F(t + \Delta t) - F(t)}{\Delta t} \leq \lim_{\Delta t \to 0} (f(t) + \varepsilon).$$
But $\lim_{\Delta t \to 0} (f(t) - \varepsilon) = \lim_{\Delta t \to 0} (f(t) + \varepsilon) = f(t),$ so $$\lim_{\Delta t \to 0} \frac{F(t + \Delta t) - F(t)}{\Delta t} = f(t),$$ and guess what, $\lim_{\Delta t \to 0} \frac{F(t + \Delta t) - F(t)}{\Delta t}$ is the definition of $F'(t)$, so we have shown that $$f(t) = F'(t).$$
That is why the difference between the values of a function at two points, $F(b) - F(a)$ is the area under the graph of the derivative of $F(x)$ between $x=a$ and $x=b$. That is, the height of the curve, which is the derivative of $F(x)$, tells us how fast the area to the left of $x=b$ would increase if we started increasing $b$. A more "tangible" interpretation is that if you place a ruler vertically across the graph, that is, the ruler is parallel to the $y$ axis and to the right of the $y$ axis, and you then start pulling the ruler to the right (keeping it parallel to the $y$ axis the whole time), the instantaneous rate at which the area under the curve to the left of the ruler is growing at any moment is the height of the graph at that moment.
By suitably defining what we mean by "area under the curve" and by other mechanisms, we can extend this idea to deal with what happens on the left side of the $y$ axis, to deal with curves that dip below the $x$ axis, and to deal with the graphs of discontinuous functions (with some limitations; not every function can be integrated).
$\endgroup$ $\begingroup$Perhaps it would help to think of a an example from real life. Imagine position. Position is an integral of velocity. If I want to find a change in position, what I do its subtract position 2 from position 1. But what does position 1 mean on it's own? Nothing, it is just telling me how far I am from an arbitrary reference point. But if I have 2 points I can begin to talk about the distance between them, which in the case of a graph corresponds to area under the derative curve. Evaluating the integral function at a specific point doesn't really mean anything without another point to put it in context.
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