The symbol ∫ was invented by Leibniz to represent the integral. It is a stretched-out S, from the Latin word for sum. This symbol is a powerful reminder of the whole construction: Sum approaches integral, S approaches ∫, and rectangular area approaches curved area:
curved area = ∫v(x) dx = ∫√x dx. (1)
The rectangles of base Δx lead to this limit—the integral of √x.The "dx" indicates that Ax approaches zero. The heights vj of the rectangles are the heights v(x) of the curve. The sum of vj times Δx approaches "the integral of v of x dx." You can imagine an infinitely thin rectangle above every point, instead of ordinary rectangles above special points.
We now find the area under the square root curve. The "limits of integration" are 0 and 4. The lower limit is x = 0, where the area begins. (The start could be any point x = a.) The upper limit is x = 4, since we stop after four years. (The finish could be any point x = b.) The area of the rectangles is a sum of base Δx times heights √x. The curved area is the limit of this sum. That limit is the integral of √x from 0 to 4:
limΔ→0[(√Δx)(Δx) + (√2Δx)(Δx) + … + (√4)(Δx)] = ∫x=0x=4√x dx. (2)
The outstanding problem of integral calculus is still to be solved. What is this limiting area? We have a symbol for the answer, involving ∫ and √x and dx—but we don't have a number.
I wish I knew who discovered the area under the graph of √x.It may have been Newton. The answer was available earlier, but the key idea was shared by Newton and Leibniz. They understood the parallels between sums and integrals, and between differences and derivatives. I can give the answer, by following that analogy. I can't give the proof (yet)—it is the Fundamental Theorem of Calculus.
In algebra the difference ƒj − ƒj-1 is vj. When we add, the sum of the v's is ƒn − ƒ0. In calculus the derivative of ƒ(x) is v(x). When we integrate, the area under the v(x) curve is ƒ(x) minus ƒ(0). Our problem asks for the area out to x = 4:
5B (Discrete vs. continuous, rectangles vs. curved areas, addition vs. integration)
The integral of v(x) is the difference in ƒ(x):
If dƒ/dx = √x then area = ∫x=0x=4√x dx = ƒ(4) − ƒ(0). (3)
What is ƒ(x)? Instead of the derivative of √x, we need its "antiderivative." We have to find a function ƒ(x) whose derivative is √x. It is the opposite of Chapters 2-4, and requires us to work backwards. The derivative of xn is nxn-1 —now we need the antiderivative. The quick formula is ƒ(x) = xn+1/(n+1) —we aim to understand it.
Solution Since the derivative lowers the exponent, the antiderivative raises it. We go from x1/2 to x3/2. But then the derivative is (3/2)x1/2. It contains an unwanted factor 3/2. To cancel that factor, put 2/3 into the antiderivative:
ƒ(x) = (2/3)x3/2 has the required derivative V(X)= x1/2 = √x.
There you see the key to integrals: Work backward from derivatives (and adjust).
Now comes a number—the exact area. At x = 4 we find x3/2 = 8. Multiply by 2/3 to get 16/3. Then subtract ƒ(0) = 0:
| ∫ | x=4 | √x dx = | 2 | (4)3/2 − | 2 | (0)3/2 = | 2 | (8) = | 16 | . (4) |
| x=0 | 3 | 3 | 3 | 3 |
The total income over four years is 16/3 = 5(1/3) million dollars. This is ƒ(4) − ƒ(0). The sum from thousands of rectangles was slowly approaching this exact area 5(1/3).
Other areas The income in the first year, at x = 1, is (2/3)(1)3/2 = 2/3 million dollars. (The false income was 1 million dollars.) The total income after x years is (3/2)x3/2, which is the antiderivative ƒ(x). The square root curve covers 2/3 of the overall rectangle it sits in. The rectangle goes out to x and up to √x, with area x3/2, and 2/3 of that rectangle is below the curve. (1/3 is above.)
Other antiderivatives The derivative of x⁵ is 5x⁴. Therefore the antiderivative of x⁴ is x⁵/5. Divide by 5 (or n+1) to cancel the 5 (or n+1) from the derivative. And don't allow n+1 = 0:
The derivative v(x) = xn has the antiderivative ƒ(x) = xn+1 / (n+1).
EXAMPLE 1 The antiderivative of x² is (1/3)x³. This is the area under the parabola v(x) = x². The area out to x = 1 is (1/3)(1)³ − (1/3)(0)³, or 1/3.
Remark on √x and x² The 2/3 from √x and the 1/3 from x² add to 1. Those are the areas below and above the √x curve, in the corner of Figure 5.3. If you turn the curve by 90°, it becomes the parabola. The functions y = √x and x = y² are inverses! The areas for these inverse functions add to a square of area 1.
You already know the area of a triangle. The region is below the diagonal line v = x in Figure 5.4. The base is 4, the height is 4, and the area is ½(4)(4) = 8. Integration is not required! But if you allow calculus to repeat that answer, and build up the integral ƒ(x) = ½x² as the limiting area of many rectangles, you will have the beginning of something important.
The four rectangles have area 1 + 2 + 3 + 4 = 10. That is greater than 8, because the triangle is inside. 10 is a first approximation to the triangular area 8, and to improve it we need more rectangles.
The next rectangles will be thinner, of width Δx = 1/2 instead of the original Δx = 1. There will be eight rectangles instead of four. They extend above the line, so the answer is still too high. The new heights are 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4. The total area in Figure 5.4b is the sum of the base Δx = 1/2 times those heights:
area = ½(1/2 + 1 + 3/2 + 2 + … + 4) = 9 (which is closer to 8).
Question What is the area of 16 rectangles? Their heights are ¼, ½, …, 4.
Answer With base Δx = ¼ the area is ¼(¼ + ½ + … + 4) = 8½.
The effort of doing the addition is increasing. A formula for the sums is needed, and will be established soon. (The next answer would be 8¼.) But more important than the formula is the idea. We are carrying out a Iimiting process, one step at a time. The area of the rectangles is approaching the area of the triangle, as Δx decreases. The same limiting process will apply to other areas, in which the region is much more complicated. Therefore we pause to comment on what is important.
What requirements are imposed on those thinner and thinner rectangles? It is not essential that they all have the same width. And it is not required that they cover the triangle completely. The rectangles could lie below the curve. The limiting answer will still be 8, even if the widths Δx are unequal and the rectangles fit inside the triangle or across it. We only impose two rules:
The area under the graph is defined to be the limit of these rectangular areas, if that limit exists. For the straight line, the limit does exist and equals 8. That limit is independent of the particular widths and heights—as we absolutely insist it should be.
Section 5.5 allows any continuous v(x). The question will be the same—Does the limit exist? The answer will be the same—Yes. That limit will be the integral of v(x), and it will be the area under the curve. It will be ƒ(x).
EXAMPLE 2 The triangular area from 0 to x is ½(base)(height) = ½(x)(x). That is ƒ(x) = ½x². Its derivative is v(x) = x. But notice that ½x² + 1 has the same derivative. So does ƒ = ½x² + C, for any constant C. There is a "constant of integration" in ƒ(x), which is wiped out in its derivative v(x).
EXAMPLE 3 Suppose the velocity is decreasing: v(x) = 4 − x. If we sample v at x = 1,2,3,4, the rectangles lie under the graph. Because v is decreasing, the right end of each interval gives vmin. Then the rectangular area 3 + 2 + 1 + 0 = 6 is less than the exact area 8. The rectangles are inside the triangle, and eight rectangles with base ½ come closer:
rectangular area = ½(3½ + 3 + … + ½ + 0) = 7.
Sixteen rectangles would have area 7½. We repeat that the rectangles need not have the same widths Δx, but it makes these calculations easier.
What is the area out to an arbitrary point (like x = 3 or x = 1)? We could insert rectangles, but the Fundamental Theorem offers a faster way. Any antiderivative of 4 − x will give the area. We look for a function whose derivative is 4 − x. The derivative of 4x is 4, the derivative of ½x² is x, so work backward:
to achieve dƒ/dx = 4 − x choose ƒ(x) = 4x − ½x².
Calculus skips past the rectangles and computes ƒ(3) = 7½. The area between x = 1 and x = 3 is the dference 7½ − 3½ = 4. In Figure 5.5, this is the area of the trapezoid.
The ƒ-curve flattens out when the v-curve touches zero. No new area is being added.
We have to distinguish two different kinds of integrals. They both use the antiderivative ƒ(x). The definite one involves the limits 0 and 4, the indefinite one doesn't:
The definite integral is definitely 8. But the indefinite integral is not necessarily 4x − ½x². We can change ƒ(x) by a constant without changing its derivative (since the derivative of a constant is zero). The following functions are also antiderivatives:
ƒ(x) = 4x − ½x² + 1, ƒ(x) = 4x − ½x² − 9, ƒ(x) = 4x − ½x² + C.
The first two are particular examples. The last is the general case. The constant C can be anything (including zero), to give all functions with the required derivative. The theory of calculus will show that there are no others. The indefinite integral is the most general antiderivative (with no limits):
indefinite integral ƒ(x) = ∫v(x) dx = 4x − ½x² + C. (5)
By contrast, the definite integral is a number. It contains no arbitrary constant C. More that that, it contains no variable x. The definite integral is determined by the function v(x) and the limits of integration (also known as the endpoints).It is the area under the graph between those endpoints.
To see the relation of indefinite to definite, answer this question: What is the definite integral between x = 1 and x = 3? The indefinite integral gives ƒ(3) = 7½ + C and ƒ(1) = 3½ + C. To find the area between the limits, subtractf at one limit from f at the other limit:
| ∫ | 3 | v(x)dx = ƒ(3) − ƒ(1) = (7½ + C) − (3½ + C) = 4. (6) |
| x=1 |
The constant cancels itself! The definite integral is the diflerence between the values of the indefinite integral. C disappears in the subtraction.
The difference ƒ(3) − ƒ(1) is like ƒn &minu; ƒ0. The sum of vj from 1 to n has become "the integral of v(x) from 1 to 3." Section 5.3 computes other areas from sums, and 5.4 computes many more from antiderivatives. Then we come back to the definite integral and the Fundamental Theorem:
| ∫ | b | v(x)dx = | ∫ | b | dƒ | dx = ƒ(b) − ƒ(a). (7) |
| a | a | dx |
