Integral under the curve

Figure 8 shows Pj. versus p for RAj and RA polymerisations. The Integrals under the curves, which have to be evaluated numerically, give Pr e = 0.0604 for f=3 and 0.1306 for f=4. Comparison with the experimentally derived values in Table I gives the results summarised... 

Figure 8. Number of gel-gel reacting pairs per initial number of groups (Pj.) versus extent of reaction (p) in RA and RA polymerisations. The integrals under the curves give...

The slope of the curves in Fig. 2.11, from day 10 onwards, corresponds to a field-loss rate of 0.058 d-1 (compare Table 2.16). Assuming this continues indefinitely, the integrals under the curves are 0.27 m2d l-1 for 90Sr and 2.2 m2d l-1 for 137Cs, values of km which are compared in Table 2.19 with those calculated previously. 

N. G. Parsonage (Imperial College, London) If one integrates under the curve for a = 0.0007 in Figure 3, one obtains 60 kcal/mole for the excess heat. This seems to be much too large for a process concerned with physical sorption and suggests that a chemical process is involved. At these temperatures on the sieve, a chemical reaction would appear to be quite possible. Would you agree ... 

A normal distribution (Eq. (9.2)) in which = 0. The X axis is scaled by the standard deviation, a, and the y axis is scaled hy a The shaded area is the integral under the curve for plus or minus one standard deviation,... 

Air Rate Air rate through the cake, and thus vacuum pump capacity, can be determined from measurements of the air flow for various lengths of dry time. Figure 18-118 represents instantaneous air rate data. The total volume of gas passing through the cake during a dry period is determined by integrating under the curve. 

The first question one may ask is What is the total mass in stars obtained by integrating under the curve in Figure 33 The data points in Figure 33 were derived assuming a Salpeter IMF (with slope —2.35) between M = 100 Mq and 0.1 M0. However, for a... 

Fig. A.7 Relation between the increase of the antiderivative F(x) (above) and the area (integral) under the curve dF/dx =f(x) (below) (a, b) Approximate, (c, d) After the limiting process.

FIGURE 135.5 Normalized time-resolved spectra of wild-type (WT) and D96N protein, both obtained at pH 7. Formation and decay times are shown, as well as the integral under the curve. Note the time is registered in... 

The integral under the heat capacity curve is an energy (or enthalpy as the case may be) and is more or less independent of the details of the model. The quasi-chemical treatment improved the heat capacity curve, making it sharper and narrower than the mean-field result, but it still remained finite at the critical point. Further improvements were made by Bethe with a second approximation, and by Kirkwood (1938). Figure A2.5.21 compares the various theoretical calculations [6]. These modifications lead to somewhat lower values of the critical temperature, which could be related to a flattening of the coexistence curve. Moreover, and perhaps more important, they show that a short-range order persists to higher temperatures, as it must because of the preference for unlike pairs the excess heat capacity shows a discontinuity, but it does not drop to zero as mean-field theories predict. Unfortunately these improvements are still analytic and in the vicinity of the critical point still yield a parabolic coexistence curve and a finite heat capacity just as the mean-field treatments do. 

A Iraditional or one-dimensional integral corresponds to the area under the curve between Ihc imposed limit, as illustrated in Figure 1.11. Multiple integrals are simply extensions of llu vc ideas to more dimensions. We shall illustrate the principles using a frmction of two vai ialiles,/(r. yj. The double integral... 

Evaluation of a one-dimensional integral using the trapezium rule. The area under the curve is approximated mm of the areas of the trapeziums. 

Fig. 8.2 Simple Monte Carlo integration, (a) The shaded area under the irregular curve equals the ratio of the number of random points under the curve to the total number of points, multiplied by the area of the bounding area, (b) An estimate of tt can be obtained by generating random numbers within the square, v then equals the number of points within the circle divided by the total number of points within the square, multiplied by 4.

Along with the curve fitting process, TableCurve also calculates the area under the curve. According to the previous discussion, this is the entropy of the test substance, lead. To find the integral, click on the numeric at the left of the desktop and find 65.06 as the area under the curve over the range of x. The literature value depends slightly on the source one value (CRC Handbook of Chemistry and Physics) is 64.8 J K mol. ... 

The program in Problem 4 gives final values for the integral under the normal curve that are obviously too large. The last entr y is 0.5002, whereas, from the nature of the problem, we know that the integral cannot exceed 0.5000. Suggest a reason for this. 

Note that /4 = 0 when capillary condensation is complete.) Integration by measurement of the area under the curve of ln(p°/p) against n between the stated limits therefore gives the value of A, which is the area of the walls of the cores, not of the pores (cf. Fig. 3.28). 

AUC is the area under the curve or the integral of the plasma levels from zero to infinite time. Conversely, equation 1 may be used to calculate input rates of dmg that would produce steady-state plasma levels that correspond to the occurrence of minor or major side effects of the dmg. 

Coulometry. If it can be assumed that kinetic nuances in the solution are unimportant and that destmction of the sample is not a problem, then the simplest action may be to apply a potential to a working electrode having a surface area of several cm and wait until the current decays to zero. The potential should be sufficiently removed from the EP of the analyte, ie, about 200 mV, that the electrolysis of an interferent is avoided. The integral under the current vs time curve is a charge equal to nFCl, where n is the number of electrons needed to electrolyze the molecule, C is the concentration of the analyte, 1 is the volume of the solution, and F is the Faraday constant. 

By using vapor-liquid equilibrium data the above integral can be evaluated numerically. A graphical method is also possible, where a plot of l/(y - xj versus Xr is prepared and the area under the curve over the limits between the initial and fmal mole fraction is determined. However, for special cases the integration can be done analytically. If pressure is constant, the temperature change in the still is small, and the vapor-liquid equilibrium values (K-values, defined as K=y/x for each component) are independent from composition, integration of the Rayleigh equation yields ... 

Radiation heat flux is graphically represented as a function of time in Figure 8.3. The total amount of radiation heat from a surface can be found by integration of the radiation heat flux over the time of flame propagation, that is, the area under the curve. This result is probably an overstatement of realistic values, because the flame will probably not bum as a closed front. Instead, it will consist of several plumes which might reach heights in excess of those assumed in the model but will nevertheless probably produce less flame radiation. Moreover, the flame will not bum as a plane surface but more in the shape of a horseshoe. Finally, wind will have a considerable influence on flame shape and cloud position. None of these eflects has been taken into account. 

Thus, the technique consists of a transformation from the time differential dt to the area differential dQ, and the essential effect of this transformation is a reduction by one of the apparent order of the reaction. The variable 6 is the area under the curve of Cb vs. time from t = 0 to time t. With modem computer techniques for integrating experimental curves, this method should be attractive. 

Graphical integration shows the area under the curve. Figure 8-38A, to be 15.764. Appl)dng this to ... 

From Figure 9-72 the area under the curve, y versus (1 y)m/(l y) (y y ) is only slightly larger than the y versus 1/y - y for this case. To avoid confusion the figure was only integrated for the latter. However, it could be performed for the former and the result should be very close to 5.89. 

A measure of the actual amount of drug in the body can be obtained from the area under the curve of the temporal concentration curve (calculated by integration). Interestingly, the temporal behavior of a drug can be extremely important in therapeutics. For example, consider three preparations of a drug that present identical values for area under the curve (i.e., amount of drug absorbed) but have different kinetics of absorption (Figure 8.23). As shown, preparation B produces a useful profile whereby the concentration exceeds the minimal effective concentration... 

Equation (2.18) is another example of a line integral, demonstrating that 6q is not an exact differential. To calculate q, one must know the heat capacity as a function of temperature. If one graphs C against T as shown in Figure 2.8, the area under the curve is q. The dependence of C upon T is determined by the path followed. The calculation of q thus requires that we specify the path. Heat is often calculated for an isobaric or an isochoric process in which the heat capacity is represented as Cp or Cy, respectively. If molar quantities are involved, the heat capacities are C/)m or CY.m. Isobaric heat capacities are more... 

Solution A graph of Cp,m against T is shown in Figure 2.9. The area under the curve is q. It can be obtained by integrating equation (2.18) with Cp,m given by equation (2.19)... 

Figure A1.3 The area between A and B under the curves (a), (b), and (c) gives the value for the line integrals. Since the areas are different, the values of the line integrals are different and depend upon the path.

Two simple numerical methods are often used to determine the area under the curve that equals the desired integral. They involve the use of the trapezoidal rule and Simpson s rule. 

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