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$$\frac{d}{dx}\int_a^xf(t)\,dt$$

I would love to to understand what exactly is the point of FTC. I'm not interested in mechanically churning out solutions to problems. It doesn't state anything that isn't already known. Prior to reading about FTC, the integral is defined as the anti-derivative. So, it's basically an operator. "Take the anti-derivative by figuring out whose derivative this is!" Simple. So, what is so "fundamental" about redundantly restating the very definition of the integral? (The derivative of the anti-derivative is the function). This to me is like saying $-(-1) = +1$. Not exactly earth shattering.

Am I missing something with regard to the indefinite vs. definite integral?

If we look at a simple example, $$\frac{d}{dx}\int_1^xt^2 \, dt = \cdots =x^2$$

Can we discuss what exactly this is representing?

  1. Why would you even write this? Why would you take the rate of change of an area under the curve? Why would you want to take the derivative of an integral? Or, is this just done to prove something else? When would you even come across this situation in Math? Taking the rate of change of the area under a curve and/or total displacement? (derivative of the definite integral)
  2. Also, what is the significance of using $t$ as a variable?
  3. Why would you integrate from a constant to a function in the first place? (take area under the curve or compute total displacement)

I don't understand what exactly things FTC even allows anyone to do. Without FTC, I can already evaluate definite integrals. Without FTC, I can already take derivatives. So, with FTC, I can take an integral then take a derivative? So, what's even the point of FTC? I really don't see anything "fundamental" whatsoever about this redundant self-evident "theorem". This is like taking the inverse of an inverse. Right back to f(x), but that's simply a "neat trick" vs. a "Fundamental Theorem of Algebra".

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    $\begingroup$ The definition of (one kind of) an integral is the limit of a riemann sum, how do I know a priori that the given limit is an antiderivative? $\endgroup$ – Tyler Dec 11 '14 at 0:12
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    $\begingroup$ You should sue whoever taught you that the definition of the integral is an antiderivative. $\endgroup$ – Bruno Joyal Dec 11 '14 at 0:13
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    $\begingroup$ As for "without FTC I can evaluate integrals," how do you personally evaluate definite integrals without it? Do you compute a limit of Riemann sums every single time? $\endgroup$ – Nick D. Dec 11 '14 at 0:19
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    $\begingroup$ @NickD. My guess would be that it appears obvious to the OP because he's been taught that an integral is an antiderivative. $\endgroup$ – David Dec 11 '14 at 0:23
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    $\begingroup$ @JackOfAll You're asking too many questions in the same post (which sounds quite rant-ish as well); narrow down a specific question for better feedback. As for why one would differentiate the "area under a curve," think of physically relevant quantities that are related by integration, e.g. velocity vs. distance traveled. Sometimes you are only given an integral as the definition of one quantity, and it's useful to know how to differentiate to get the other. $\endgroup$ – Gyu Eun Lee Dec 11 '14 at 0:30

13 Answers 13

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I am guessing that you have been taught that an integral is an antiderivative, and in these terms your complaint is completely justified: this makes the FTC a triviality.

However the "proper" definition of an integral is quite different from this and is based upon Riemann sums. Too long to explain here but there will be many references online.

Something else you might like to think about however. The way you have been taught makes it obvious that an integral is the opposite of a derivative. But then, if the integral is the opposite of a derivative, this makes it extremely non-obvious that the integral can be used to calculate areas!

Comment: to keep the real experts happy, replace "the proper definition" by "one of the proper definitions" in my second sentence.

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    $\begingroup$ Riemann is fine, but you could go even further to Lebesgue ;) $\endgroup$ – Tobias Kienzler Dec 11 '14 at 8:52
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    $\begingroup$ @Tobias Hence the final comment in my answer :) $\endgroup$ – David Dec 11 '14 at 11:32
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    $\begingroup$ @TobiasKienzler: or then back to sums with the Henstock-Kurzweil integral, which integrates all the Lebesgue-integrable functions and then some, with much easier definition. $\endgroup$ – mbork Dec 13 '14 at 21:22
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    $\begingroup$ @mbork Neat, I didn't know about that one $\endgroup$ – Tobias Kienzler Dec 14 '14 at 12:36
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    $\begingroup$ I am the OP. I am truly grateful and humbled by knowledge shared by this collective brain trust. In the next few days, I am going to do this thread justice and read everything when I can be alone for a few hours next week. Thank you again for this. $\endgroup$ – JackOfAll Dec 14 '14 at 23:12
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You seem to think that you already know that definite integrals have something to do with antidifferentiation. Probably you think this because $\int_a^b f(x) \, dx$ looks remarkably similar to $\int f(x) \, dx$.

But, without the FTC, these two things have nothing whatsoever to do with one another.

They are two completely unrelated operations which for some bizarre reason share a symbol.

$\int f(x) \, dx$, as you note, means the antiderivative of $f(x)$.

But $\int_a^b f(x) \, dx$ means the area between the curve you get when you graph $f(x)$ and the $x$-axis, over the interval $[a,b]$. Without the FTC, there is no reason to expect this to have anything to do with the antiderivative (or "indefinite integral".)

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    $\begingroup$ @MackTuesday: Try that for $f(x) = \frac1x$ and see what happens. (Hint: Your definition fixes the integration constant so that $F(0) = 0$; if the antiderivative $F$ of $f$ diverges at zero, this is not possible.) Yes, of course there is a relationship between the two concepts, but then, that relationship pretty much is the FTC. $\endgroup$ – Ilmari Karonen Dec 11 '14 at 17:03
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    $\begingroup$ @IlmariKaronen ...and yet the point remains that they're not unrelated. Note that without bounds, the antiderivative sign also represents the "indefinite integral"--so of course it's related to the definite integral! $\endgroup$ – Kyle Strand Dec 12 '14 at 0:55
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    $\begingroup$ The bizarre reason is precisely the fundamental theorem of calculus :) $\endgroup$ – Roberto Bonvallet Dec 12 '14 at 17:55
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    $\begingroup$ @KyleStrand, there's no "of course the definite integral and the indefinite integral are related", unless you know the FTC. That was Kundor's point. Despite the relationship of the symbols (or the names), they have completely different origins (definitions). People have created these symbols and names because they were able to prove the FTC. $\endgroup$ – Paul Draper Dec 14 '14 at 4:33
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    $\begingroup$ I am the OP. I am truly grateful and humbled by knowledge shared by this collective brain trust. In the next few days, I am going to do this thread justice and read everything when I can be alone for a few hours next week. Thank you again for this. $\endgroup$ – JackOfAll Dec 14 '14 at 23:13
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As integrals and derivatives are presented in Apostol's Calculus, it becomes quite evident that the relationship between them--the Fundamental Theorem of Calculus--is quite remarkable and a bit unexpected.

Apostol actually introduces the notion of an integral first: the notation $$\int_{x=a}^b f(x) \, dx$$ is intended to represent the signed area enclosed by a function $f(x)$ and the $x$-axis, on the interval $x \in [a,b]$. This idea of "area" is something familiar to us from elementary geometry, and it is not difficult to conceptualize the "area under a curve" as an extension of the areas of more familiar geometric shapes, such as polygons and circles. Thus it seems natural to talk about the area enclosed by the curve of a parabola $f(x) = x^2$ and the $x$-axis on the interval $[0,1]$. Indeed, Archimedes of Syracuse, thousands of years ago, used a method remarkably similar to Riemann sums to obtain areas enclosed by parabolic segments.

Now let's switch gears and talk about derivatives: a derivative $f'(x)$ of a function $f(x)$ at a point $x = a$ has the geometric interpretation of the slope of the tangent line to the function at that point. Loosely speaking, the greater this value, the more rapidly the function $f(x)$ is increasing at that point. More formally, $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}.$$

What makes the integral and derivative concepts in calculus (or analysis, if you prefer), is that both are mathematical ideas involving some kind of limiting process: the (Riemann) integral is understood as the sum of the rectangular areas defined by successively more refined partitions of the interval $[a,b]$, and the derivative is understood as the slope of a secant line as one intersection point approaches the other.

Note that in these contexts, it is not at all obvious that the two concepts are related. Yet the Fundamental Theorem of Calculus states (in one form) that $$\int_{x=a}^b f(x) \, dx = F(b) - F(a)$$ where $F(x)$ is some function satisfying $F'(x) = f(x)$. This gives us a means to compute without resorting to Riemann summation a definite integral as the difference of the integrand's antiderivative on the interval's endpoints. (In fact, the FTC is a unidimensional special case of Stokes' Theorem and as such holds deeper insights, but that's not in the scope of our discussion).

So, in summary, the FTC is not a trivial result. Apostol does in fact provide a quasi-geometric heuristic "proof" of why this relationship should exist, and it is worth reading. And if we are to have a proper appreciation for calculus, it helps to have the proper pedagogy and motivation that his text provides. But should you desire to understand the foundations of calculus further, then a more rigorous and less computationally oriented treatment is recommended, such that found in Walter Rudin's Principles of Mathematical Analysis.

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  • $\begingroup$ You clarified it for me, thanks! $\endgroup$ – PatrickT Dec 12 '14 at 17:24
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    $\begingroup$ Ah, so F(b) - F(a) is a form of that theorem? I've been taught that F(b) - F(a) is just "the Newton-Leibniz formula"... $\endgroup$ – myfreeweb Dec 13 '14 at 19:23
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    $\begingroup$ @myfreeweb Some mathematicians actually DEFINE the integral by the Newton-Liebniz formula.While this may be pedagogically much easier for beginning calculus students to understand then Riemann sums or the even more sophisticated formulations of the integal, it's a very mathematically misleading thing to do to say the least. $\endgroup$ – Mathemagician1234 Dec 14 '14 at 5:47
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    $\begingroup$ I am the OP. I am truly grateful and humbled by knowledge shared by this collective brain trust. In the next few days, I am going to do this thread justice and read everything when I can be alone for a few hours next week. Thank you again for this. $\endgroup$ – JackOfAll Dec 14 '14 at 23:13
  • $\begingroup$ This made the point the most clearly. The integral is actually defined independently of the derivative. The integral is merely defined the anti-derivative, but it is the Reimann sum of rectangles under the curve. Later on, one that can attempt to tie them together as "inverses using the FTC. $\endgroup$ – JackOfAll Aug 26 '15 at 19:24
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From an intuitive standpoint, $F(x)=\int_a^xf(t)dt$ can be viewed as a cumulative function that tallies up the values of $f$ from $a$ to whatever $x$ is. With this in mind, it shouldn't be surprising that $\frac{d}{dx}F(x)=f(x)$.

From a theoretical standpoint, FTC part 2 is the theorem that allows us to write $$\int_a^b f(t)dt=\left.F(t)\right|_a^b$$ where $F(t)$ is an antiderivative of $f(t)$. In other words, FTC2 allows us to evaluate definite integrals using indefinite integrals.

FTC allows us to define new functions by integrating others, such as $$\operatorname{erf}(x)\triangleq\int_2^x e^{t^2}dt\hspace{30pt}\hspace{30pt}\operatorname{Li}(x)\triangleq\int_2^x \frac{dt}{\operatorname{ln}(t)}$$ $$\operatorname{C}(x)\triangleq\int_2^x \cos{t^2}dt\hspace{20pt}\hspace{30pt}\operatorname{S}(x)\triangleq\int_2^x \sin{t^2}dt$$

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The fundamental theorem of calculus is just a continuous generalization of telescoping series.

Suppose you have a sequence of numbers, $$x_1,~x_2,~x_3,~\dots,~x_n,$$ like, for example, $1,2,5,7,12$.

You can consider the sequence of differences between each number and the next one, $$x_2-x_1,~x_3-x_2,~x_4-x_3,~\dots,~x_n-x_{n-1},$$ which in the example would be, $1,3,2,5$.

If you add up the differences, most of the terms in the sum cancel and you get the total difference between the first and the last number, \begin{align} & (x_2-x_1) + (x_3-x_2) + (x_4-x_3) + \dots +(x_n-x_{n-1}) \\ &= -x_1 + (x_2 - x_2) + (x_3 - x_3) + \dots + (x_{n-1}-x_{n-1}) + x_n \\ &= x_n - x_1. \end{align} In the example this is, $1 + 3 + 2 + 5 = 11 = 12-1$.

In the fundamental theorem of calculus the concept is the same, but with the following replacements:

  • Instead of a sequences you have functions.
  • Instead of differences you have the derivative.
  • Instead of sums you have the integral.
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    $\begingroup$ This explains the theorem, not the point of the theorem. $\endgroup$ – Jessica B Dec 11 '14 at 7:16
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    $\begingroup$ The point is that it extends basic ideas about discrete things (sums, differences) to the continuous setting... $\endgroup$ – Nick Alger Dec 11 '14 at 7:19
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    $\begingroup$ Many of the top-voted answers say that the relationship between integrals and derivatives is "surprising". But I feel it is quite intuitive, as this answer demonstrates. Admittedly we have to prove that the idea still works as we take smaller and smaller slices - presumably that is what the FTC proves. But why do they say it is surprising? $\endgroup$ – joeytwiddle Dec 13 '14 at 8:54
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The importance of the fundamental theorem of calculus- and some of the other posters have given correct responses, don't get me wrong- can be best understood in a historical context that goes back to a century before Riemann constructed his precise definition of the integral.

The basic idea of the integral is essentially that of areas and volumes, which dates back to Ancient Greece and Archimedes' "method of exhaustion," by which these quantities were computed by inscribing polygons of known area into an arbitrary region with smaller and smaller areas until the region is "filled up" and then adding up all the areas. (If this sounds like basically the definition of Riemann sums, except with more general figures then rectangular partitions- well, you're right. In a lot of ways, taking the limit of a Riemann sum is a rigorous formulation of exhaustion.) As you could well imagine, this was an incredibly cumbersome and lengthy procedure for any but the simplest figures. For example, it took Archimedes months to obtain a reasonable approximation of the area of a circle using this method. Another, much simpler example was Archimedes' use of the procedure to obtain the area under a parabola, which was fairly important in not only mathematics but physics and construction problems. A good discussion of the details can be found here. You can get the idea from these two examples that using this procedure meant that something we usually take for granted from calculus as a relatively simple computation was a Herculean task fit only for geniuses to undertake.

Later mathematicians, from the Renaissance onward, used geometric methods and calculations with limits to obtain areas and volumes. For example, Galileo was able the guess the value of the area of one arch of the cycloid, a curve generated by a rolling circle in the plane,to be $3\pi^2$. A good discussion of the details of how a geometric proof of this fact goes- comparing it to a calculus solution- can be found here. Again, this was somewhat easier, but not much.

The relationship between the tangent and the quadrature problems was first recognized by Issac Newton's teacher, Issac Barrow, and fully exploited by Newton, Liebniz, and the Bernoullis. Since most functions that were known from both basic geometry and physics at that point were fairly smooth and had antiderivatives, the new science of calculus, unified by the Fundamental Theorem, made areas and volumes now a relatively straightforward computation to solve for all mathematics and science students. Without it, it's hard to imagine that calculus- and for that matter, most of classical mechanics and subsequent breakthroughs in physics and other hard sciences- would have been possible. Furthermore, if it were possible, it would have taken dozens of centuries to achieve. Also, since most of the later developments in the theory of calculus-such as the Riemann integral and its subsequent refinements, as well as developments in differential equations and functional analysis, were all abstracted largely from calculus, none of these things would have been likely, either. So you can make a very good case calculus, and its descendant, analysis, would have died stillborn without some version of the Fundamental Theorem of Calculus.

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There are some cute applications of the fundamental theorem of calculus, and I'm sure some of the others answers will dig them up. But for the most part I agree with you: the FTC, in one dimension, isn't all that exciting! But it's just the tip of the iceberg.

You have focused on one form of the fundamental theorem of calculus; there is a second form, namely that

$$\int_a^b \frac{df}{dx}\,dx = f(b) - f(a)\tag{1} $$

I wouldn't blame you for thinking this is even more obvious than the first form! But the FTC is one special case, and the most boring one at that, of a general principal called Stokes's Theorem that is much more deserving of the "fundamental" moniker.

Let's take a closer look at what equation 1 is saying. The left-hand side asks you to take some (possibly horrible and complicated) function $f$, take its derivative, and then sum the values of that derivative up over the entire interval $[a,b]$. FTC says you can get the same answer just by looking at $f$ at two values. The key points, here, are that

  1. The LHS requires knowing $f$ (and its derivative) everywhere along $[a,b]$. The RHS needs $f$ only at the boundary of the interval.
  2. The LHS requires being able to take the derivative of $f$. The right-hand side requires only knowing $f$, and no derivatives.

Well duh, you're thinking. Isn't that the entire point of anti-derivatives. Yes, indeed. But it turns out that both of these benefits carry over, in a beautiful way, to higher dimensions, where an equivalent property holds. Let's say you have a region of the plane $\Omega$, and its boundary curve $\partial \Omega$. Then:

$$\int_{\Omega} \nabla \cdot v \,dV = \int_{\partial \Omega} v\cdot \hat{n}\,dA.$$

There is some fairly elementary intuition about what these terms mean, but going into it without at least a bit of vector calculus knowledge would take us too far astray... the key point though is that the above lets you turn integrals over areas in the plane to integrals over their one-dimensional boundaries, just like the FTC turns a one-dimensional integration into a zero-dimensional difference of values. It doesn't always work, but it works enough of the time to be extremely powerful.

For example, let's say you draw a closed curve $\gamma(s)$ in the plane. What is the area enclosed by the curve? Maybe you've learned some tools for computing this area: some slicing techniques, perhaps. You've also seen that these techniques are a huge pain; even moreso when the curve $\gamma$ is complicated with lots of loops and concavities. It turns out you can compute the area enclosed by only integrating around $\gamma$:

$$\textrm{Area} = \int \frac{1}{2} \gamma(s) \cdot \gamma'(s)^{\perp}\, ds.$$

The two-dimensional problem has become a one-dimensional problem, and much more tractable both analytically and computationally. You can find similar nice formulas for many other geometric quantities of interest, such as the center of mass of a region $\Omega$, its moment of inertia, etc -- all quantities that nominally depend on the entire interior of $\Omega$ -- using only integration around the boundary.

One final example: let's say I give you a point $p$ in the plane, and a super-complicated closed curve $\gamma$. How can you tell if the point is inside, or outside, the region enclosed by $\gamma$? People use various tricks to do this, for example by drawing a ray from $p$ to a point at infinity, and counting how many times the ray intersects $\gamma$... but you can do it robustly and easily using a boundary integral,

$$\int \frac{-1}{4\pi\|\gamma(s) - p\|^2}[\gamma(s)-p]\cdot \gamma'(s)^{\perp}\,ds$$

which will be equal to $1$ if $p$ is inside the region, or $0$ if $p$ is outside (assuming I've not made any mistakes in my calculation).

There's more: recall the second key point about the FTC above: the LHS requires computing derivatives, while the RHS does not. This comes up all of the time when doing numerical calculations and simulations. For example, let's say you want to simulate the way that your clothing wrinkles and folds as you dance. It turns out that if you represent your shirt as a surface parameterized by $r(s,t): \mathbb{R}^2 \to \mathbb{R}^3$, the bending energy of the shirt, which is required to correctly compute shirt physics, is given by $$E \propto \int (\Delta r \cdot \hat{n})^2\,dA.$$ Again I won't go too much into the details of the math; the important part is that computing $\Delta r$ requires knowing two derivatives of $r$. This means that $r$ must be twice-differentiable for the formula to make sense; that's fine in an ideal setting, but what if the geometry of your shirt comes from a Microsoft Kinect, or is inferred form video footage? The shirt surface will be "chunky," or have lots of noise, and you often won't even be able to compute first derivatives, never mind second derivatives. It turns out that Stokes's Theorem can be used to reduce the number of derivatives that are needed, and is behind the cloth animation you see in video games and movies effects. In fact, countless physical simulations, from how the galaxy formed, to how wind flows around an airplane wing, to how your cheek deforms when you get punched in the face, rests on a foundation made up of the "pointless trick" that is the (generalized) FTC.

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    $\begingroup$ "the FTC, in one dimension, isn't all that exciting!" I think the past 500 years of mathematics would disagree with that statement. $\endgroup$ – Daniel McLaury Dec 11 '14 at 6:19
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    $\begingroup$ So the fact that knowing the forces acting on a moving body allows you to determine its trajectory isn't important to you? What kind of applied math do you do, exactly? $\endgroup$ – Daniel McLaury Dec 11 '14 at 6:34
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    $\begingroup$ "Second the practice of approximating an integral curve of a Hamiltonian system is an exercise in purely definite integration." Yes, and this is true because of the fundamental theorem of calculus! $\endgroup$ – Daniel McLaury Dec 11 '14 at 6:45
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    $\begingroup$ Why are people downvoting this answer?-- it seems a bit antiintellectual to do so; it's the only one that hints at the enlightening, abstract notion of what an integral means... and best points the way to further learning. You may disagree about whether the 1-dimension FTC is exiting, but the author clearly indicated that this was an opinion. It's a little disappointing to see all the other answers fixate on the 1-dimensional case, since there's so much more beauty and meaning to the FTC. $\endgroup$ – Max Wallace Dec 12 '14 at 22:00
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    $\begingroup$ In my opinion, a good answer entails some inference about the mathematical background of the individual asking the question, and furnishes a response in that context. A student who asks the question that was asked is highly unlikely to be in a position to appreciate Stokes' Theorem. That's not what was asked. And for what it's worth, I don't think the one-dimensional case is trivial: that would imply no need for rigorous proof. If someone asks you about the quadratic formula, you don't respond with a treatise on Galois theory and solvability by radicals. $\endgroup$ – heropup Dec 13 '14 at 8:21
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$$\frac{d}{dx}\int_a^xf(t)\,dt\tag{1}$$

  1. Why would you even write this? Why would you take the rate of change of an area under the curve? Why would you want to take the derivative of an integral? Or, is this just done to prove something else? When would you even come across this situation in Math? Taking the rate of change of the area under a curve and/or total displacement? (derivative of the definite integral)
  2. Also, what is the significance of using $t$ as a variable?
  3. Why would you integrate from a constant to a function in the first place? (take area under the curve or compute total displacement)

A lot going on here and plenty of good answers already. But I'll chime in question by question.

Question 1

We might write $(1)$ because we are confronted by a function which is defined in terms of an integral such as $$A(x)=\int_a^x f(t)\,dt\tag{2}$$ that we want to "do" calculus on just like we "did" calculus on a multitude of other functions (polynomials, exponentials, logs, trig, products, quotients, compositions, etc.): find instantaneous rates of change, linear approximations, local and/or global extrema, intervals of increase/decrease, intervals of concavity, inflection points, etc. In order to do those things, we want to get our hands on $A'(x)$. The FTC offers an extremely efficient way to do so.

To get a geometric feel for functions of the form of $(2)$, consider $f(t)=\sin t$ (in blue) and $g(x)=\int_0^x \sin t\,dt$ in red.

enter image description here

As $x$ varies here from $x=0$ to $x=2\pi$, the different amounts of area under the curve $y=f(t)$ are accumulated indicated by the red shading. This "accumulator function" $g(x)$ is itself a bona fide function. At each fixed $x$, we can compute the accumulated area under $f$ from $0$ to that $x$, resulting in the number $g(x)$. We can then plot the point $(x,g(x))$ and repeat the process to generate the graph of $g(x)$ in red.

Hopefully once you see that functions like $g(x)$ make sense, then it is natural to want to do calculus on them.


Question 2

In $(1)$, $t$ is a dummy variable of integration that ranges from $a$ to $x$. If we wrote $$\int_a^x f(x)\,dx,$$ this is a different animal than $(1)$: the $x$ in the integrand here is varying from the lower limit of integration $a$ to the upper limit of integration $x$. Is this what we want to convey? Likely not (at this level). The notation in $(1)$ using a different symbol for the dummy variable of integration than the independent variable in the upper limit of integration brings clarity and precision to what it is that we want to communicate.


Question 3

For example, if $v(t)$ is a velocity function for an object on $a\le t\le b$ then $d(t):=\int_a^t v(s)\,ds$ represents the (net) displacement of the object over from time $a$ to time $t$. Note that $d(t)$ is indeed a function of the independent variable time $t$ and it is very natural to want to do calculus on $d(t)$, e.g., take its derivative with respect to $t$.

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You do not seem to notice that there is a problem: If $f$ is any function on $[a,b]$, then there is an important question:

Is there a differentiable function $F$ on $[a,b]$ such that $F'=f$ ?

Obviously if there is one such fuction, then there are many (add something constant). But is there such a thing in the first place??

The symbol $$ \int_a^x f(t) dt $$

is not a new notation for such a function (if one exists at all). This thing might give you a real number for every $x\in [a,b]$ and this works for any continous $f$ by a limit of sums etc. This definition (limit of sums for partitions) can be given even before one ever talks about derivatives!

So you define a new function $F$ out of a given continous $f$ by $$ F(x):=\int_a^x f(t) dt $$ and now the FTC states that $F'=f$, and this is something you have to prove.

Exercise: If you assume that $$ F(x)=\int_0^x e^{t^2} dt $$ is just a function with derivative $e^{t^2}$ please figure out the value $F(1)$

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Integration is really, really much more than an operation. What you're missing is that integration is analogous to summation, and that area is just an interpretation of it.

I think the easiest way to see that the antiderivative is the area under the curve, is with the geometrical interpretation of the funtamental thoerem of calculus on wikipedia:

enter image description here

If you think about the function $A(x)$ as the "area under the curve up to $x$", then you'll have:

$$A(x+h)-A(x) \approx f(x)\cdot h\implies \frac{A(x+h)-A(x)}{h}\approx f(x)$$ So, when $h$ tends to $0$, geometrically you have a better estimative for the area, and algebraically, you approach the derivative for the area function $A(x)$. The $f(x)$ doesn't depend on $h$, so it stays the same.

Therefore you're really saying that:

$$\lim_{h\to 0} \frac{A(x+h)-A(x)}{h} = A'(x) = f(x)$$

So the area of the function $f$ is the function such that its derivative is $f$ itself. This is the antiderivative. But what exactly is area? What the fundamental theorem of calculus will do, is that it will replace "area" by an infinite sum of little retangles, a method called Riemman Sum: enter image description here And will let their base $\Delta x$ tend to $0$, to get a better approximation of the 'area'. Then, the theorem will replace $A(x)$ by $\int f(x) dx$, there $dx$ means that $\Delta x \to 0$. The theorem will take the derivative of this more precise definition of 'area' and prove its derivative is the function itself.

But remember, the power of the integral isn't in the antiderivation process, or calculation of area. It's really related to infinite sums over continuous domain.

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  • $\begingroup$ In the following, why are you able to say A'(x) = f(x)? I missed the connection. Did you take the limit of both sides in the section prior? $$\lim_{h\to 0} \frac{A(x+h)-A(x)}{h} = A'(x) = f(x)$$ $\endgroup$ – JackOfAll Aug 26 '15 at 22:03
  • $\begingroup$ ie: Did you imply this step? $$\frac{A(x+h)-A(x)}{h}\approx f(x)$$ $$\lim_{h\to 0} \frac{A(x+h)-A(x)}{h} \approx \lim_{h\to 0} f(x)$$ $$\lim_{h\to 0} \frac{A(x+h)-A(x)}{h} \approx f(x)$$ $$A'(x) \approx \lim_{h\to 0} f(x)$$ $\endgroup$ – JackOfAll Aug 26 '15 at 22:05
  • $\begingroup$ And why did you suddenly turn the $\approx$ into $=$ ? $\endgroup$ – JackOfAll Aug 26 '15 at 22:09
  • $\begingroup$ @JackOfAll I did not gave a proof, just an intuition. What I'm saying is that $ \frac{A(x+h)-A(x)}{h}\approx f(x)$ when $h\to 0$, thus, since the derivative is just the limit of this expression, we expect $\lim_{h\to 0} \frac{A(x+h)-A(x)}{h}$ to be $f(x)$. For a formal proof, you can follow the wikipedia's page on 'fundamental theorem of calculus'. $\endgroup$ – Lucas Zanella Aug 28 '15 at 1:50
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I don't know whether the following will persuade you. One of the points of FTC is that a continuous function on an interval is the derivative of SOMETHING, and you can use this to define functions with particular desired properties. For example, the function $f: f(x) = \frac{1}{x}$ is continuous on $(0, \infty)$, so that $\int_{1}^{x} \frac{1}{t} dt$ exists for all $x > 0.$ Now you probably know the (natural) $\log$ function, as a function with familiar properties, but FTC can be used to demonstrate the existence of a function with the right properties from scratch, even if you hadn't known before that there was such a function. For if we set $g(x) = \int_{1}^{x} \frac{1}{t} dt,$ then we know that $g^{\prime}(x) = \frac{1}{x}.$ Also, a change of variables shows that $g(ab) = g(a) + g(b)$ for positive real $a$ and $b$, since $\int_{1}^{ab} \frac{1}{t} dt = \int_{1}^{a} \frac{1}{t} dt +\int_{a}^{ab} \frac{1}{t} dt,$ and setting $u = \frac{ t}{a},$ we obtain $ g(ab) = \int_{1}^{ab} \frac{1}{t} dt = \int_{1}^{a} \frac{1}{t} dt +\int_{1}^{b} \frac{1}{u} du = g(a) + g(b).$

In some books, such as Spivak's "Calculus", this approach is used to define the exponential function, rather than just using a definition using power series. The inverse function $h = g^{-1} : \mathbb{R} \to (0,\infty)$ satisfies $h^{\prime}(x) = h(x)$ for all $x$ and $h(0) =1.$ Then when the theory of Taylor series is developed, it is clear that $h$ has the familiar Taylor series.

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    $\begingroup$ It's not true that "a continuous function on an interval is the integral of SOMETHING"; I think you must have made an editing mistake. $\endgroup$ – ruakh Dec 11 '14 at 4:46
  • $\begingroup$ @ruakh : Oops, yes, absolutely: I meant " a continuous function is the DERIVATIVE of SOMETHING", which is why I went on to discuss the definition of the log function using the integral of $\frac{1}{x},$ which I still think is a useful example if you have some apppreciation of rigour. $\endgroup$ – Geoff Robinson Dec 11 '14 at 20:15
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In my perspective your question is about the Second Fundamental Theorem of Calculus.

I had your same doubt! But now I understand the power of this amazing theorem.

What is the advantage?

Some integrals are impossible to solve without a computer. And what if you are in a test and you cannot use a computer or a calculator?

Or what if you are solving a system of differential equations?

With this theorem you do not need to solve the integral. You can skip that part!

Maybe you are in Calculus I and you will not use too much, but in Calculus III and Differential Equations will be very important. Also in many Physics applications.

If the upper limit is x and the lower limit is a constant, the derivative cancels the integral and that is the answer!

Please check this link: http://beginnermathstackexchange.blogspot.com/2014/12/second-fundamental-theorem-of-calculus.html

The most important cases are when the limits of the integral are not only an x. If the upper limit is a function f(x) and the lower limit is a constant, and we are solving the derivative of the integral of g(x). The solution is g(f(x))(f`(x)))

In this link you will find a problem of the famous Calculus book written by the professors Larson and Edwards. http://beginnermathstackexchange.blogspot.com/2014/12/problem-from-calculus-book-written-by_16.html

If we have two different functions as upper and lower limit, then just apply the formula that is in this link: http://beginnermathstackexchange.blogspot.com/2014/12/second-fundamental-theorem-of-calculus_16.html

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  • $\begingroup$ You should explicitly say if that is your own blog $\endgroup$ – Aditya Hase Dec 17 '14 at 17:32
  • $\begingroup$ I just created a website to put the images. I did not find other way to do it. It is not really a blog. $\endgroup$ – Beginner Dec 17 '14 at 23:46
  • $\begingroup$ It is recommended that you type up the equations in your post using LaTeX, rather than link to images of the equations. Here is a link that explaines how to do it: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Nick Alger Dec 25 '14 at 22:52
  • $\begingroup$ I did not know about LaTex, but I found it a couple of days ago and I am trying to learn it. However, I really appreciate the link that you sent me! $\endgroup$ – Beginner Dec 26 '14 at 1:54
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The only reason you know the definition of an integral as an antiderivative is BECAUSE of the FTC.

And there are all sorts of situations where it is useful in the real world. For example, if you integrate force along a path, you get energy. Let's say that that force is gravity. FTC says that know matter whether you push a boulder straight up the side of a mountain or take a bunch of switch-backs, the rock still has the same amount of gravitational potential energy at the top of the mountain.

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