A Little Trouble With the Fundamental Theorem of Calculus.

Question

I'm currently studying calculus. I understand pretty well with the math and with the main concepts.

But there is something I don't fully understand, "The fundamental Theorem of Calculus". Not in a comprehensive way, but in a conceptual way.

I saw this two videos:

• https://youtu.be/rfG8ce4nNh0?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr
• https://youtu.be/pWtt0AvU0KA

And I completely understand the math and the proof. But it leaves me with a bad taste on my mouth, they seem completely unrelated, but they explain the same thing.

How it is posible that an area under the velocity curve gives (in the case of a car moving) its position at that current time. And how it is possible just to know how much space someone has covered just by computing this:

$$ \int^T_0{v(t) \; \mathrm d t} = r(T) - r(0) = \Delta r $$

It feels anti-intiutive. When you sum, you add the previous value and so on. $$ Area = A_i + A_{i + 1} ... A_{i + n} $$ For example, in the case of the car, you know the previous position, so you add it up. $$ r_i = r_{i-1} + v_0 \; \Delta t $$

Could someone explain me why these two concepts are related and why $\int_a^b{f(x) \; \mathrm d x} = F(b) - F(a)$ gives you the area under a curve.

Thanks :)

Answer 1

The area under the velocity function does not give "the position at that current time".

The area under the velocity function gives the total displacement: how far from the initial position taken at the initial time (the left edge) you are at the final time (the right edge).

This actually goes back all the way to Oresme. Let's consider two very basic/straightforward situations:

1. If the velocity is constant, $v_0$, then the graph is a straight horizontal line. The total displacement after a period of $a$ time units will be exactly $av_0$ distance units. On the other hand, the area under the graph from time $t_0$ to time $t_1$ equals the height, $v_0$, times the length of the base, $t_1-t_0$ (which equals $a$ time units): but this is exactly $av_0$, the displacement we expect just from physical considerations.

2. If the motion is uniformly accelerated function, so that the graph of the velocity is a straight line; say the initial velocity is $v_0$, the final velocity is $v_1$, and it takes $a$ times units to get there. It is an observable physical fact that in this situation, the displacement is the average between the displacement you would have obtained using the initial velocity, and the displacement you would get using the final velocity: that is, $\frac{1}{2}(av_0+av_1)$ distance units. On the other hand, the area under the graph consists of a rectangle of height $v_0$ and width $a$, plus a triangle sitting on top of it of base $a$ and height $v_1-v_0$. The total area is then $$av_0 + \frac{1}{2}(a(v_1-v_0)) = \frac{1}{2}av_1 + \frac{1}{2}av_0 = \frac{1}{2}(av_0+av_1).$$ Same as the displacement calculation.

One of the contents of the First Part of the Fundamental Theorem of Calculus, especially if you view the integrand as a velocity function, is that these two observations extend from uniformly accelerated motion to arbitrary continuous motion: for any continuous velocity function, the area under the velocity function will be the same as the displacement caused by that motion.

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As to why the integral and the derivative seem to be connected when they are defined so disparately... yes, this should be a surprise! The fact of the Fundamental Theorem of Calculus and that they seem to be kind of inverses of each other is surprising. It's what makes the FTC such an important piece of mathematics, even beyond its applicability in calculations of areas. There is no a priori reason why you might expect a calculation of slopes of tangents to be connected with a measurement of areas... and yet they are. I treat it as a wonderful surprise, and not as something that one might necessarily expect. This is especially true, IMHO, of the Second Part of the theorem.

Answer 2

$\int_a^b f(x) dx = F(a) - F(b)$ is defined to be the area under the curve in the interval $[a,b].$

We break the area under the curve into rectangular regions, and sum the area of these rectangles. 3blue1brown starts discussing this at the 4 minute mark.

I hope it is obvious why these rectangles make a reasonable approximation of the area. What is less obvious is that if we partition the region into smaller and smaller rectangles the error in our approximation gets smaller and smaller. And at the limit, this error shrinks to zero.

This is the Reimann Sum definition of the integral.

Is this what is bothering you?

The fundamental theorem of calculus goes on to show how Reimann sum relate to derivatives.