When the endpoints are fixed at a and b, we have a definite integral. When the upper limit is a variable point x, we have an indefinite integral. More generally: When the endpoints depend in any way on x, the integral is a function of x. Therefore we can find its derivative. This requires the Fundamental Theorem of Calculus.
The essence of the Theorem is: Derivative of integral of v equals v. We also compute the derivative when the integral goes from a(x) to b(x)—both limits variable.
Part 2 of the Fundamental Theorem reverses the order: Integral ofderivative off equals ƒ + C. That will follow quickly from Part 1, with help from the Mean Value Theorem. It is Part 2 that we use most, since integrals are harder than derivatives.
After the proofs we go to new applications, beyond the standard problem of area under a curve. Integrals can add up rings and triangles and shells—not just rectangles. The answer can be a volume or a probability—not just an area.
Start with a continuous function v. Integrate it from a fixed point a to a variable point x. For each x, this integral ƒ(x) is a number. We do not require or expect a formula for ƒ(x)—it is the area out to the point x. It is a function of x! The Fundamental Theorem says that this area function has a derivative (another limiting process). The derivative dƒ/dx equals the original v(x).
5C (Fundamental Theorem, Part 1) Suppose v(x) is a continuous function:
if ƒ(x) = ∫ax then dƒ/dx = v(x).
The dummy variable is written as t, so we can concentrate on the limits. The val of the integral depends on the limits a and x, not on t.
To find df ldx, start with Δƒ = ƒ(x + Δx) − ƒ(x) = difference of areas:
| Δƒ = | ∫ | x+Δx | v(t) dt − | ∫ | x | v(t) dt = | ∫ | x+Δx | v(t) dt. (1) |
| a | a | x |
Officially, this is Property 1. The area out to x + Δx minus the area out to x equals the small part from x to x + Δx. Now divide by Δx:
| Δƒ | = | 1 | ∫ | x+Δx | v(t) dt = average value = v(c). (2) |
| Δx | Δx | x |
This is Property 7, the Mean Value Theorem for integrals. The average value on this short interval equals v(c). This point c is somewhere between x and x + Δx (exact position not known), and we let Δx approach zero. That squeezes c toward x, so v(c) approaches u(x)—remember that v is continuous. The limit of equation (2) is the Fundamental Theorem:
| Δƒ | → | dƒ | and v(c)→v(x) so | Δƒ | = v(x). (3) |
| Δx | dx | dx |
If Δx is negative the reasoning still holds. Why assume that v(x) is continuous? Because if v is a step function, then ƒ(x) has a corner where dƒ/dx is not v(x).
We could skip the Mean Value Theorem and simply bound v above and below:
| for t between x and x + Δx: | vmin ≤ v(t) ≤ vmax | (4) |
| integrate over that interval: | vminΔx ≤ Δƒ ≤ vmaxΔx | |
| divide by Δx: | vmin ≤ Δƒ/Δx ≤ vmax |
As Δx→0, vmin and vmax approach v(x). In the limit dƒ/dx again equals v(x).
Graphical meaning The ƒ-graph gives the area under the v-graph. The thin strip in Figure 5.14, has area Δƒ. That area is approximately v(x) times Δx. Dividing by its base, Δƒ/Δx is close to the height v(x). When Δx → 0 and the strip becomes infinitely thin, the expression "close to" converges to "equals." Then dƒ/dx is the height at v(x).
When the upper limit is x, the derivative is v(x). Suppose the lower limit is x. The integral goes from x to b, instead of a to x. When x moves, the lower limit moves. The change in area is on the left side of Figure 5.15. As x goesforward, area is removed. So there is a minus sign in the derivative of area:
| The derivative of g(x) = | ∫ | b | v(t) dt is | dg | = −v(x). (5) |
| x | dx |
The quickest proof is to reverse b and x, which reverses the sign (Property 3):
| g(x) = − | ∫ | x | v(t) dt so by Part 1 | dg | = −v(x). |
| b | dx |
The general case is messier but not much harder (it is quite useful). Suppose both limits are changing. The upper limit b(x) is not necessarily x, but it depends on x. The lower limit a(x) can also depend on x (Figure 5.15b). The area A between those limits changes as x changes, and we want dA/dx:
| If A = − | ∫ | b(x) | v(t) dt then | dA | = v(b(x)) | db | −v(a(x)) | da | . (6) |
| a(x) | dx | dx | dx |
The figure shows two thin strips, one added to the area and one subtracted.
First check the two cases we know. When a = 0 and b = x, we have daldx = 0 and db/dx = 1. The derivative according to (6) is v(x) times 1 —the Fundamental Theorem. The other case has a = x and b = constant. Then the lower limit in (6) produces −v(x). When the integral goes from a = 2x to b = x³, its derivative is new:
| EXAMPLE 1 A = | ∫ | x³ | cos t dt = sin x³ − sin 2x dA/dx = (cos x³)(3x²) − (cos 2x)(2). |
| 2x |
That fits with (6), because db/dx is 3x² and da/dx is 2 (with minus sign). It also looks like the chain rule—which it is! To prove (6) we use the letters v and ƒ:
| A = | ∫ | b(x) | v(t) dt = ƒ(b(x)) − ƒ(a(x)) (by Part 2 below) |
| a(x) |
| dA | = ƒ'(b(x)) | db | − ƒ'(a(x)) | da | (by the chain rule) |
| dx | dx | dx |
Since ƒ' = v, equation (6) is proved. In the next example the area turns out to be constant, although it seems to depend on x. Note that v(t) = 1/t so v(3x) = 1/3x.
| EXAMPLE 2 A = | ∫ | 3x | 1 | dt has | dA | = (1/3x)(3) − (1/2x)(2) = 0. |
| 2x | t | dx |
| Question A = | ∫ | x | v(t) dt has | dA | = v(x) + v(−x). Why does v(−x) have a plus sign? |
| -x | dx |
We have used a hundred times the Theorem that is now to be proved. It is the key to integration. "The integral of dƒ/dx is ƒ(x) + C." The application starts with v(x). We search for an ƒ(x) with this derivative. If dƒ/dx = v(x), the Theorem says that
| ∫ | v(x) dx = | ∫ | dƒ | dx = ƒ(x) + C. |
| dx |
We can't rely on knowing formulas for v and ƒ only the definitions of ∫ and d/dx.
The proof rests on one extremely special case: dƒ/dx is the zero function. We easily find ƒ(x) = constant. The problem is to prove that there are no other possibilities: ƒ' must be constant. When the slope is zero, the graph must be flat. Everybody knows this is true, but intuition is not the same as proof.
Assume that dƒ/dx = 0 in an interval. If ƒ(x) is not constant, there are points where ƒ(a) ≠ ƒ(b). By the Mean Value Theorem, there is a point c where
| ƒ'(c) = | ƒ(b)−ƒ(a) | (this is not zero because ƒ(a) ≠ ƒ(b)). |
| b−a |
But ƒ'(c) ≠ 0 directly contradicts dƒ/dx = 0. Therefore ƒ(x) must be constant.
Note the crucial role of the Mean Value Theorem. A local hypothesis (dƒ/dx = 0 at each point) yields a global conclusion (ƒ = constant in the whole interval). The derivative narrows the field of view, the integral widens it. The Mean Value Theorem connects instantaneous to average, local to global, points to intervals. This special case (the zero function) applies when A(x) and ƒ(x) have the same derivative:
If dA/dx = dƒ/dx on an interval, then A(x) = ƒ(x) + C. (7)
Reason: The derivative of A(x) − ƒ(x) is zero. So A(x) − ƒ(x) must be constant.
Now comes the big theorem. It assumes that v(x) is continuous, and integrates using ƒ(x):
5D (Fundamental Theorem, Part 2) If v(x) = dƒ/dx then ∫ab v(x) dx = ƒ(b) − ƒ(a).
Proof The antiderivative is ƒ(x). But Part 1gave another antiderivative for the same v(x). It was the integral—constructed from rectangles and now called A(x):
| A(x) = | ∫ | x | v(t) dt also has | dA | = v(x). |
| a | dx |
Since A' = v and ƒ' = v, the special case in equation (7) states that A(x) = ƒ(x)+ C. That is the essential point: The integral from rectangles equals ƒ(x) + C.
At the lower limit the area integral is A = 0. So ƒ(a)+ C = 0. At the upper limit ƒ(b) + C = A(b). Subtract to find A(b), the definite integral:
| A(b) = | ∫ | b | v(x) dx = ƒ(b) − ƒ(a). |
| a |
Calculus is beautiful—its Fundamental Theorem is also its most useful theorem.
Another proof of Part 2 starts with ƒ' = v and looks at subintervals:
The left sides add to ƒ(b) − ƒ(a). The sum on the right, as Δx → 0, is ∫ab v(x) dx.
Up to now the integral has been the area under a curve. There are many other applications, quite different from areas. Whenever addition becomes "continuous," we have integrals instead of sums. Chapter 8 has space to develop more applications, but four examples can be given immediately—which will make the point.
We stay with geometric problems, rather than launching into physics or engineering or biology or economics. All those will come. The goal here is to take a first step away from rectangles.
EXAMPLE 3 (for circles) The area A and circumference C are related by dA/dr = C.
The question is why. The area is πr². Its derivative 2πr is the circumference. By the Fundamental Theorem, the integral of C is A. What is missing is the geometrical reason. Certainly πr² is the integral of 2nπ, but what is the real explanation for A = ∫ C(r) dr?
My point is that the pieces are not rectangles. We could squeeze rectangles under a circular curve, but their heights would have nothing to do with C. Our intuition has to take a completely different direction, and add up the thin rings in Figure 5.16.
Suppose the ring thickness is Δr. Then the ring area is close to C times Δr. This is precisely the kind of approximation we need, because its error is of higher order (Δr)². The integral adds ring areas just as it added rectangular areas:
| A = | ∫ | r | C dr = | ∫ | r | 2πr dr = πr². |
| 0 | 0 |
That is our first step toward freedom, away from rectangles to rings.
The ring area ΔA can be checked exactly—it is the difference of circles:
ΔA = π(r + Δr)² − πr² = 2πr Δr + π(Δr)².
This is CΔr plus a correction. Dividing both sides by Δr → 0 leaves dA/dr = C.
Finally there is a geometrical reason. The ring unwinds into a thin strip. Its width is Δr and its length is close to C. The inside and outside circles have different perimeters, so this is not a true rectangle—but the area is near CΔr.
EXAMPLE 4 For a sphere, surface area and volume satisfy A = dV/dr.
What worked for circles will work for spheres. The thin rings become thin shells. A shell goes from radius r to radius r + Δr, so its thickness is Ar. We want the volume of the shell, but we don't need it exactly. The surface area is 4πr² , so the volume is about 4πr² Δr. That is close enough!
Again we are correct except for (Δr)². Infinitesimally speaking dV = A dr:
| V = | ∫ | r | A dr = | ∫ | r | 4πr² dr = ¾πr³. |
| 0 | 0 |
This is the volume of a sphere. The derivative of V is A, and the shells explain why. Main point: Integration is not restricted to rectangles.
EXAMPLE 5 The distance around a square is 4s. Why does the area have dA/ds = 2s?
The side is s and the area is s². Its derivative 2s goes only half way around the square. I tried to understand that by drawing a figure. Normally this works, but in the figure dA/ds looks like 4s. Something is wrong. The bell is ringing so I leave this as an exercise.
EXAMPLE 6 Find the area under v(x)= cos-1 x from x = 0 to x = 1.
That is a conventional problem, but we have no antiderivative for cos-1 x. We could look harder, and find one. However there is another solution-unconventional but correct. The region can be filled with horizontal rectangles (not vertical rectangles). Figure 5.17b shows a typical strip of length x = cos v (the curve has v = cos-1 x). As the thickness Δv approaches zero, the total area becomes ∫ x dv. We are integrating upward, so the limits are on v not on x:
| area = | ∫ | π/2 | cos v dv = sin v | π/2 | = 1. | |
| 0 | 0 |
The exercises ask you to set up other integrals—not always with rectangles. Archimedes used triangles instead of rings to find the area of a circle.
