Is the integral of secx convergent or divergent between 0 to π/2?
I myself asked this question but I have got my answer and if anybody has a doubt can refer it...Actually secx at x=π/2 becomes undefined or ∞ is obtained And as given in the question the limit is from 0--->π/2 then the integral becomes a improper one and when this integral is solved by putting Lim b-->π/2 and solving it we get the final answer as ∞ and thus the integral becomes "Divergent".
Is this integral divergent or convergent?
Determine whether the integral is divergent or convergent. If it is convergent, evaluate it and enter that value as your answer. If it diverges to infinity, state your answer as "INF" (without the quotation marks). If it diverges to negative infinity, state your answer as "MINF". If it diverges without being infinity or negative infinity, state your answer as "DIV". ∫ (3((x+3)^2)-4) dx [ inf on top -inf on bottom] Thank you :)
How is the integral from 0 to 1 of 3/x^5 divergent?
Integral (0 to 1, 3/x^5 dx ) Since the function is not defined at 0, this is an improper integral, and we must express this in terms of a limit. lim Integral (t to 1, 3/x^5 dx ) t -> 0+ We express this as t approaches 0 from the right because our bounds for integration go from 0 to 1, and 0+ would fall in that range. (0- would mean negative numbers). lim Integral (t to 1, 3x^(-5) dx ) t -> 0+ Solving this as normal, lim ( 3(-1/4)x^(-4) ) {evaluated from t to 1} t -> 0+ lim ( (-3/4)x^(-4) ) {evaluated from t to 1} t -> 0+ lim ( (-3/4)(1)^(-4) - (-3/4)t^(-4) ) t -> 0+ lim ( (-3/4) - (-3/4)t^(-4) ) t -> 0+ lim ( (-3/4) - (-3/4)(1/t^4) ) t -> 0+ As t approaches 0+, the behavior of 1/t^4 is such that t gets really really large (i.e. if we progressively try smaller values of t for 1/t^4, we get a large number). This behavior alone tells us that the limit approaches infinity.
What are divergent integrals?
In single-variable calculus, the definite integral of a continuous function over a closed, finite interval [math][a,b][/math] is proven to exist—these are called proper integrals.Potential problems arise when the conditions of this theorem fail, e.g., when:the function is undefined or discontinuous at some point(s) of the interval of integration, orthe interval of integration is unbounded (such as [math]\int_{-\infty}^a\cdots[/math] or [math]\int_a^{\infty}\cdots[/math].We deal with such improper integrals by first splitting them up into simple pieces (in which the trouble occurs just at one endpoint) and set up each piece as a limit of proper integrals. Whenever limits are involved like this, there is the potential for a limit not to exist, which is what the term divergence refers to: an improper integral could converge or diverge, while the value of a proper integral always exists (it, itself, is defined via limits or sup/inf, but that isn’t our focus right now).For example:[math]\int_0^{\infty}\frac1{x^2}dx[/math]is split up as, say:[math]\int_0^1\frac1{x^2}dx+\int_1^{\infty}\frac1{x^2}dx[/math].Now, the first integral only has trouble at zero, and the second only because of the infinity. We can set each up as a limit of proper integrals:[math]\displaystyle\int_0^1\frac1{x^2}dx=\lim_{\varepsilon\to0^+}\int_{\epsilon}^1\frac1{x^2}dx[/math], and[math]\displaystyle\int_1^{\infty}\frac1{x^2}dx=\lim_{M\to+\infty}\int_1^M\frac1{x^2}dx[/math].Working these inside-out (as always), we can compute each of the integrals inside the limits very easily via antiderivatives, and the values of those integrals we’re taking the limits of always exist (because they’re proper integrals).But if we compute those and take the limits, we find that the first limit does not exist (because the integrals [math]\to+\infty[/math]) and the second limit comes out to one. So, this first improper integral diverges, and the second one converges (and thus our original integral diverges, because one bad apple ruins the whole bunch here).
Why is the integral of a total divergence zero?
An alternative model to QFTs that addresses this question comes from mandalic geometry. This uses a very different language and notation. Also a quite different mathematical input is employed but arrives at a similar solution for integration of total divergence over space-time - that is the total divergence integrates to zero.However in mandalic geometry “zeros” or rather “zero alternatives” occur in every subdivision of the whole. Wherever a Cartesian ordered triplet includes one or more zeros a similar number of zero alternatives occur and contribute to the total action,A simple diagram demonstrating principles involved is shown below. The key to the hexagram structures can be found at the blog Mandalic Geometry, In brief, the icons are used to simulate a discrete hybrid 6D/3D Hilbert space made commensurate with E3 and Cartesian coordinates by the simple mathematical operation of dimensional compositing.Shown here are the vertices and origin point of a single hyperface of an n6-hypercube. The midpoints of the four edges are not shown. Each of the four hosts two hexagrams. This mandalic patterning of hexagrams unifies space and time through a new number system, Probable numbers which are a time expansion of Cartesian coordinates (R3) and ordinary Euclidean space (E3).The Probable number system is more robust than the Complex number system used in quantum field theories. It suggests a possible path to unification of QFT and GR. From there it may be just a step or two to quantum gravity, an emergent phenomenon of all the forces, not itself a single force.
How do convergent and divergent integrals differ?
I’m assuming you are studying improper integrals. So, the difference between convergent integrals and divergent integrals is that convergent integrals, when evaluated, go to a specific value whereas a divergent integral, when evaluated does not go to a finite value and goes to [math]\pm \infty[/math]. These of course represent areas. Remember that improper integrals are caused due to vertical or horizontal asymptotes being inside the bounds.
How can I regularize divergent proper integral?
If you have something like [math]f(x) = 1/x[/math] and integration on an interval containing [math]x = 0[/math] then you can exclude a small interval [math](-\epsilon, +\epsilon)[/math] around [math]x = 0[/math] and later take limits, letting [math]\epsilon \to 0[/math].
Calculus: convergent/divergent integrals?
well to determine whether the integral is convergent or divergent...you need to evaluate the function itself at the limits of integration. for 1) 4/(x-6)^3 at x = 6 is 4/(6-6)^3 = 4/0 = infinite and at x=8 4/(8-6)^3 = 0.5..since one limit is infinite this is a divergent integral. for 2) at y=0 1/(0-1) = -1 and at y =1 1/(4-1) = 1/3...so its convergent. for 3) at x=0 ln(0)/sqrt(0) = UNDEFINED....So we need L'Hopitals rule, which takes the derivative of the numerator and denominator and re-evaluates the limit in question. Using L'H rule (1/x) / (0.5)x^(-0.5) = 2x^0.5/x and when x -> 0 this goes to infinitity. So this is divergent as well. I'll let you evaluate number 2.
Is the integral of e^(-sqrt(x)) / sqrt(x) from 1 to infinity convergent or divergent?
First integrate the term let u = sqrt(x) du = dx/ [2sqrt(x)] 2du = dx/sqrt(x) evaluate the substituted upper and lower boundaries when x = 1, u = sqrt(1) = 1 when x approaches infinity, u approaches infinity INT: 2e^-u du from 1 to infinity = -2e^-u from 1 to infinity = -2e^[-sqrt(infinity)] - (-2e^ -sqrt(1)) = -2/e^sqrt(infinity) + 2/e the first term will arbitrarily become smaller and smaller and approaches 0 therefore the integral coverges to 2/e