xdx 8. d.√ a2 + x2 xdx (x2 + a2)}} d.(a2 + x2)1 = 1 × ( x2 + a2)− × 2xdx √ x2 + a2 1 d.a(a2 + x2)———‡a × (a2 + x2)−−1 × 2xdx ——a (a2 + x2)− × 2xdx The term employed for this process is to differentiate, differentiating, or differentiation, according to the connection in which the term is used. Now, the reverse process-that is, from the differential to obtain the quantity from which the differential is supposed to have been or may be derived-is to integrate the differential, called, also, integrating or integration. The sign or symbol used to indicate this process is. As the sign ✔ to denote root is derived from the sign or letter R, or ✔, so, as the sign for integration, is derived from a form of the letter S, or, meaning the sum of all the increments, or, in fluxions, the sum of all possible positions of the flowing quantity. To integrate a differential, reverse the process of differentiating; that is, increase the index denoting the power by unity, divide by the index so increased, and also by the differential of the root. Examples.-1. Taking the preceding examples, and reversing the 5. S(2axdx + 12x2dx + 20x3dx) = √2axdx + §12x2dx + √20x3dx ax2 + 4x3 + 5xa. 6. S(2bydy + 3dy)=f2bydy + Sзdy =by2 + 3y + C.* * It will be seen that in the reverse process of example 6 we do not obtain the known quantities a and -4, annexed by the sign + or —, which were in the function from which this differential was obtained. These, being constant quantities, have no differential; and, being connected with the variable terms by the signs plus and minus, they Definition. A quantity to differentiate or to integrate is called a function. Example.-If we have a function of the form u 5x3 + 2x2 + 6x, we have du 15x2dx+4xdx + 6dx = (15x2 + 4x + 6)dx. Hence du 15x2 + 4x + 6. Now, this last quantity (15x2 + 4x + 6) is called the differential co-efficient of the function u, because it is what the differential, da, is multiplied by in the value of du. And, du as is what this differential co-efficient is equal to always, the dx du symbol is adopted as denoting the differential co-efficient. Hence dx the differential co-efficient of any function is the differential of the function divided by the differential of the variable. This quotient expresses, also, the ratio of the differential of the function to the differential of the variable. The limit of this ratio, which is of much use in investigations by the calculus, is determined by supposing the variable, as x in the preceding example, to become du 0. Then we have (that is, the differential of the function u divided by the differential of the variable x) 15x2 + 4x + 6, which, necessarily disappear in the differential, and consequently they cannot be restored in the integration, because the differential would be unaffected no matter how many of these constants were in the quantity differentiated connected by the signs + and nor of what value they were. The whole sum of these constants are, therefore, in integrations, represented by the letter C, implying the sum of these constants, as in example 6. In fact, C is always to be understood as appended to the integral of any function, whether expressed or not. In many problems in the calculus the value of C can be determined by the conditions of the problem. when x= 0, becomes 6, and this is the limiting ratio of the differential of the function 5x+2x2 + 6x divided by the differential of the variable x. But, when x becomes 0, u = 5x3 + 2x2 + 6x becomes du 0 dx 0 0, and du becomes 0, and dx becomes 0, and 15x+4x+6; and much severe criticism has been bestowed upon a mode of investigation in science, and especially in mathematics, which employs such seemingly intangible nonentities. A learned author has called them the "ghosts of departed quantities," a value remaining in their relation when the quantities themselves have vanished. We must remember, however, that it is not the quantities themselves, but their ratio, which is the object to be determined, and the ratio remains unaltered through whatever changes of proportional increase or diminution the terms of the ratio may undergo. To illustrate : The ratio of 3 to 6 is Now, if each term of this ratio be multiplied by a million, the ratio is still. And if each term of the ratio is multiplied by a million for a million of times, or a million million of times, the ratio still remains unaltered, being, no more, no less. Again, if, instead of multiplying, the terms be each divided by a million, the ratio remains. And if each term of the ratio be divided by a million for a million of times, or a million million of times, the ratio continues to be, no more, no less. And that which thus maintains its fixed value, eternally, through all conceivable changes of magnitude of its terms possible, large and small, it is reasonable to conclude will retain the same value when the terms 0 each become 0, which is the value of * 0 * is the symbol for indefiniteness in quantities, and it has sometimes been used to 0 perplex young algebraists, as seeming to lead to incongruous results. To illustrate : Put xa, a being any number whatever. Then x2 a2. Take the first equation from the second, and we have x2 — x = α2 — α, or, by transposition, a-x= 112 — x2. Divide by a-x, and we have 1 =a+x=2a. 2a. Now, taking a=1, 14, 2, 21, 3, 31, etc., we have 1 2 3 4 5 7, etc. to infinity. ་་་་ 6 a, Again, put x=== a, as before, then x3 = a3; subtracting, x3x=a3a, or a― x = a3 x3. Divide by ax, and we have 1 = a2 + a + ï2 = 3, 2, where a may be any value from 1 to infinity. In like manner a— x — α4 —— x4. a4 x4. Divide by a-x, and 1= a3 + a2x + ax2+x3=4a3, where a may be any value from 1 to infinity. This process, if continued till the powers of x and a are infinite, will give, in each case, an infinite number of values for the quotient of the first term, α х It will be observed, however, that in all these cases, since xa, x — a — α=0, x2 = 0, x2 — a2 PROBLEM.-When y' represents a line or area which moves continually parallel to itself, and its value in any position is known, either definitely or in terms of x, then the value of the surface or solid, generated by such quantity in its movement, will be fy'da, being the distance through which the line or surface moves. Simpson's Fluxions, or Young's Integral Calculus.) (See Example 1.-To find the area of a parallelogram whose base and height are given. A Put the height AG a, the base CD=b, AH, and y- EF, which is supposed to begin to move at CD, and to move parallel to itself up the line GA to AB, thus generating the parallelogram ABCD. Now, EF in this B Ꮳ case is known definitely, being equal to b=y'. ABCD = Sy'dx = Sbda=bx = (when a = to the product of the base by the altitude. Hence the area a) ab, or the area is equal NOTE.—If x be taken=ļa, we get the area of ABCD; if x=1α, we get the area, etc. Example 2.-To find the area of a triangle when the base and height are given. Here the variable line EF, or y', is supposed to move from BC parallel to itself, terminated in every position by the lines AB, AC of the triangle, and thus generating the triangle ABC. Put AD: a, BC —b, AG—x, EF—y; then a a:b::x:y is a symbol, the number of values of which, as shown above, is infinity multiplied by 0 infinity, or infinity's square. Hence is appropriately termed the symbol of indefinite ness. The same principle is familiar in practical arithmetic. Bearing in mind that the product of two factors divided by one factor gives the other factor, taking the product 0X1 0, and dividing by the factor 0, we have 1= . Also, 0x9 0; 0 0 hence 9 and so on through all area of the triangle half the product of the base by the altitude, or half the parallelogram of the same base and altitude. E A G F Example 3.-To find the area of a parabola. Here EF is the generating line for the whole parabola, beginning at CD, moving parallel to itself, and terminating at A; and EG is the generating line for the semiparabola CEAB. Put AB-a, BC=b, ᎪᏀ AG=x, and EG=y; then, by the property of the parabola (see Prob. I., "Parabola," p. 253), we have a:x:: b2: y'; whence y2 — C B D dx = S = xsdx bx3 Va (when x a) Hence the area of the whole parabola × √x × dx X CAD AB × CD the rectangle under AB and CD. Example 4.—To find the area of a semicircle whose radius is given. Here the variable line EF, or y, is supposed to move from AB, parallel to itself, limited by the extremity E of the constant line CE AB, thus generating the semicircle ADB. Now, while EF generates the semicircle ADB, EG generates, in like manner, the quadrant ADC. To find the area of the quadrant, therefore, put AC or CE =r, r, and CG=x; then EG=√ r2 — x2 y-y'. And the area of the quadrant ADC = §y'dx = §√r2 = Sy'dx = √ √ r2 — x2 × dx = § (r2 — x2)}dæ S S = (by developing (x2-2) into a series, and multiplying each term by |