# The integral of F (positive function) from A to B is the area below the graph and above the x axis.
It is calculated by approximating it with rectangles in increasing numbers. On the right, the approximation of integral of # I F is not always positive, then the areas of rectangles are signed: positive/negative if they are above/below the x axis: in the case shown below on the left, the integral between -1 and 1 of f1 is # In the case shown below on the right, the integral between -2 and 2 is 0. |
Il calcolo degli integrali di qualunque funzione può essere effettuato anche con questa calcolatrice online.
• At the beginning the integral of x → √(x³+1) is calculated.
function F(x) { with(Math) { return sqrt(pow(x,3)+1) }}
I can take 52.049971104. It is more than enough for each application.
With WolframAlpha, with "integrate sqrt(x^3+1) dx between 1 and 7" I can have:
52.049971103763749933450986697949208...
• If I change F in the following way, I can find the integral of x → exp(x²) from 0 to 3:
function F(x) { with(Math) { return exp(x*x) }}
1444.5451228899435 if a=0 b=3 n=256e4 [8.363031156477518e-9] 1444.5451228815804 if a=0 b=3 n=128e4 [3.336322151881177e-8] 1444.5451228482172 if a=0 b=3 n=64e4 [1.3355474948184565e-7] 1444.5451227146625 if a=0 b=3 n=32e4 [5.341280484572053e-7] 1444.5451221805344 if a=0 b=3 n=16e4 [0.0000021365635802794713] 1444.5451200439708 if a=0 b=3 n=8e4 [0.000008546221806682297] 1444.545111497749 if a=0 b=3 n=4e4 [0.00003418488427087141] 1444.5450773128648 if a=0 b=3 n=2e4 [0.00013673952230419672] 1444.5449405733425 if a=0 b=3 n=1e4 [1444.5449405733425]
I can take 1444.54512289. With WolframAlpha I can have: 1444.54512289271415471376001340596...
• If I change F in the following way, I can find the integral of x → sin(x²), from 0 to π²:
function F(x) { with(Math) { return sin(x*x) }}
0.6773089372424593 if a=0 b=9.8696044010893586 n=64e4 [-5.866825913969365e-10] 0.6773089378291419 if a=0 b=9.8696044010893586 n=32e4 [-2.3466496523738555e-9] 0.6773089401757916 if a=0 b=9.8696044010893586 n=16e4 [-9.386749710849074e-9] 0.6773089495625413 if a=0 b=9.8696044010893586 n=8e4 [-3.754700705904668e-8] 0.6773089871095483 if a=0 b=9.8696044010893586 n=4e4 [-1.501883707399898e-7] 0.6773091372979191 if a=0 b=9.8696044010893586 n=2e4 [-6.007597528334685e-7]
I can take 0 .677308937. With WolframAlpha I can have: 0.677308937046889033108516222836...
• If I change F in the following way, I can find the integral of x → 1/log(x), from e to 10:
function F(x) { with(Math) { return 1/log(x) }}
4.270481688429555 if a=2.7182818284590452 b=10 n=64e4 [5.799805080641818e-12] 4.2704816884237555 if a=2.7182818284590452 b=10 n=32e4 [2.254640918408768e-11] 4.270481688401209 if a=2.7182818284590452 b=10 n=16e4 [9.033929160295884e-11] 4.27048168831087 if a=2.7182818284590452 b=10 n=8e4 [3.6146730053587817e-10] 4.2704816879494025 if a=2.7182818284590452 b=10 n=4e4 [1.4457457453431743e-9] 4.270481686503657 if a=2.7182818284590452 b=10 n=2e4 [5.783186374230809e-9] 4.27048168072047 if a=2.7182818284590452 b=10 n=1e4 [4.27048168072047]
I can take 4.27048168843. With WolframAlpha I can have: 4.2704816884313611820564608183...
• The previous integrals, like most integrals, can only be calculated numerically.
But it is better to calculate numerically also many of the other integrals, also to better master the precision of the results.
An example: to calculate (in a certain sea) the kg of salt present in a column of water 1 square meter wide and 100 meters deep you have to calculate
the integral between 1 and 100 of
function F(x) { with(Math) { return 34.7*(1-0.0176*exp(-0.05*x)*(1+0.04*x)) }}
3448.491419883378 if a=0 b=100 n=8e4 [-9.836185199674218e-10] 3448.4914198843617 if a=0 b=100 n=4e4 [-4.0872691897675395e-9] 3448.491419888449 if a=0 b=100 n=2e4 [-1.637590685277246e-8] 3448.491419904825 if a=0 b=100 n=1e4 [3448.491419904825]
The value is 3448.49141988, or 3448.49. The answer (3448.50) is almost right, but ... the data is approximate. Assuming that − 0.05 and 0.04 are correct, we would have that the solution is between 3443.4 and 3453.6!!! Unfortunately, in this manual (and many others) the calculations are done in this way.
# 34.75*(1-0.01755*exp(-0.05*x)*(1+0.04*x)) 3453.5216196594883 if a=0 b=100 n=16e4 # 34.65*(1-0.01765*exp(-0.05*x)*(1+0.04*x)) 3443.4613961986424 if a=0 b=100 n=16e4
• If I change F in the following way,
I can find the integral of x →
function F(x) { with(Math) { return 1/sqrt(pow(x,3)+1) }} 2.604364224936649 if a=0 b=100 n=16e5 [-7.549516567451064e-15] 2.6043642249366568 if a=0 b=100 n=8e5 [-1.1590728377086634e-13] 2.6043642249367727 if a=0 b=100 n=4e5 [-5.950795411990839e-14] 2.604364224936832 if a=0 b=100 n=2e5 [-4.583000645652646e-13] 2.6043642249372905 if a=0 b=100 n=1e5 [2.6043642249372905] - - - - - 0.17999997958874092 if a=100 b=10000 n=1e5 [0.17999997958874092] - - - - - 0.017999999387442685 if a=10000 b=1000000 n=1e5 [0.017999999387442685]
2.604364224936649+0.18·(1+0.1+0.01+...) = 2.604364224936649 + 0.18·10/9 = 2.604364224936649 + 0.2 = 2.804364224936649 = (rounding) 2.804364224937.
• If I change F in the following way,
I can find the integral of x →
function F(x) { with(Math) { return x/(pow(x,2)+1) }} 2.30756025884133 if a=0 b=10 n=1e5 - - - - - 2.2976599247434124 if a=10 b=100 n=1e5 - - - - - 2.3025355951596063 if a=100 b=1000 n=1e5 - - - - - 2.302584597660238 if a=1000 b=10000 n=1e5
The integral between two successive powers of 10 tends to be the same positive number; I deduce that the integral diverges to infinity
• If I change F in the following way,
I can find the integral of x →
function F(x) { with(Math) { if(x==0) {u=1} else {u=sin(x)/x}; return u }} 1.892166140734187 if a=-1 b=1 n=64e5 [-2.631228568361621e-13] 1.89216614073445 if a=-1 b=1 n=32e5 [-6.661338147750939e-16] 1.8921661407344508 if a=-1 b=1 n=16e5 [-9.192646643896296e-14] 1.8921661407345427 if a=-1 b=1 n=8e5 [-4.1233683134578314e-13] 1.892166140734955 if a=-1 b=1 n=4e5 [-1.8680612612342884e-12] 1.892166140736823 if a=-1 b=1 n=2e5 [-7.574163518597743e-12] 1.8921661407443973 if a=-1 b=1 n=1e5 [1.8921661407443973]
I take 1.89216614073445 (only up to this point the variation between the outputs is divided approximately by 4).
• I want to calculate the volume of the solid pictured below.
The solid can be approximated by the union of many disks of thickness dx and radius
function F(x) { with(Math) { return PI*pow(x,4) }} 0.6283185307179291 if a=0 b=1 n=128e5 [-1.4210854715202004e-14] 0.6283185307179433 if a=0 b=1 n=64e5 [3.752553823233029e-14] 0.6283185307179058 if a=0 b=1 n=32e5 [1.6986412276764895e-13] 0.6283185307177359 if a=0 b=1 n=16e5 [5.965228311310966e-13] 0.6283185307171394 if a=0 b=1 n=8e5 [2.457922754217634e-12] 0.6283185307146815 if a=0 b=1 n=4e5 [9.810374734797733e-12] 0.6283185307048711 if a=0 b=1 n=2e5 [3.9273806429207525e-11] 0.6283185306655973 if a=0 b=1 n=1e5 [0.6283185306655973]
I take 0.6283185307180 (only up to this point the variation between the outputs is divided approximately by 4).
We have made this simple example to illustrate the use for calculating volumes. In this case we could have easily calculated the volume directly and found that it holds π/5 = 0.6283185307179586.
•
I want to calculate
function F(x) { with(Math) { return abs(x*(x-2)) }} 2.666666666668032 if a=0 b=3 n=1e8 [1.5631940186722204e-12] 2.666666666666469 if a=0 b=3 n=1e7 [1.1231016117108084e-12] 2.666666666665346 if a=0 b=3 n=1e6 [1.2369039126269854e-10] 2.6666666665416554 if a=0 b=3 n=1e5 [1.2374167468465203e-8] 2.666666654167488 if a=0 b=3 n=1e4 [2.666666654167488]
I take 2.666 : I understand that, if the rounding errors did not intervene, the outputs would continue in this way (only up to 2.666666666666469 the variation between the outputs is divided approximately by 100; if the approximation errors did not become preponderant, the next exit would be 2.66666666666666...). 2.666 = 2+2/3 = 8/3.
•
I want to calculate
(for the graph, see here)
function F(x) { with(Math) { if (x<=PI/2) {u=2*sin(x*x)} else {u=sin(x*x)} return u }} 1.6016787712180702 if a=0 b=3 n=1024e4 [-1.828905447087692e-7] 1.601678954108615 if a=0 b=3 n=512e4 [-1.6475709685437323e-13] 1.6016789541087797 if a=0 b=3 n=256e4 [7.315614130032344e-7] 1.6016782225473667 if a=0 b=3 n=128e4 [-5.5564441936439835e-12] 1.6016782225529231 if a=0 b=3 n=64e4 [-0.000002926253118484823] 1.6016811488060416 if a=0 b=3 n=32e4 [-8.699263531752877e-11] 1.6016811488930343 if a=0 b=3 n=16e4 [-3.4807401405601013e-10] 1.6016811492411083 if a=0 b=3 n=8e4 [-1.3924066344372932e-9] 1.601681150633515 if a=0 b=3 n=4e4 [-5.56947221674875e-9] 1.6016811562029871 if a=0 b=3 n=2e4 [-2.2277982347773673e-8] 1.6016811784809695 if a=0 b=3 n=1e4 [0.0000023643901312375704]
If I wanted more figures I would take 1.601679. In our case 1.602 is sufficient.
The graph was plotted with R (see), defining function F as follows: F = function(x) ifelse(x <= pi/2, 2*sin(x*x), sin(x*x))
There are not many functions that can be formally integrated, even if (who knows why? ...) they are the only ones that can be integrated in textbooks (not only secondary school). A famous function that cannot be elementally integrated is the Gaussian. /// Non sono molte le funzioni che si sanno integrare formalmente, anche se (chissà perché? ) sono le sole che si trovano da integrare sui libri di testo (non solo di scuola secondaria). Una famosa non integrabile elementarmente è la gaussiana.
Elementary functions (functions of one variable which are the composition of arithmetic operations, powers, exponentials, logarithms, trigonometric functions and their inverses) are closed with respect to derivation but not with respect to integration. Some examples of elementary functions that do not have an elementary antiderivative, using x as the input variable: /// Le funzioni elementari (funzioni di una variabile che sono la composizione di operazioni aritmetiche, potenze, esponenziali, logaritmi, funzioni trigonometriche e loro inverse) sono chiuse rispetto alla derivazione ma non rispetto all'integrazione. Alcuni esempi di funzioni elementari che non hanno una antiderivata elementare, usando x come variabile di input:
√(1+x^3), √(1-x^4), 1/√(1+x^4), 1/log(x), log(log(x)),
exp(x^2), exp(-x^2), exp(x)/x, exp(exp(x)),