Integral (x^2) e^(-2x) from 0 to infinity?
∫x² * e^(-2x) dx, from 0 to ∞ lim b ---> ∞: ∫x² * e^(-2x) dx, from 0 to b Applying integration by part where: u = x² du = 2x dx dv = e^(-2x) v = (-1/2)*e^(-2x) lim b ---> ∞: (-1/2)x²(e^-2x) + ∫x * e^(-2x) dx Applying integration by part once again will give us: u = x du = dx dv = e^(-2x) v = (-1/2)e^(-2x) lim b ---> ∞: (-1/2)x²(e^-2x) + [ (-1/2)*x*(e^-2x) + (1/2)*∫e^(-2x) dx ] lim b ---> ∞: (-1/2)x²(e^-2x) - (1/2)*x*(e^-2x) - (1/4)(e^-2x) Plugging in the limits: lim b ---> ∞: [ (-1/2)(b)²(e^-2b) - (1/2)(b)(e^-2b) - (1/4)(e^-2b) ] - [ 0 - 0 - (1/4) ] As you can see, when we plugged in '∞' for all values of x, we got an indeterminate form of (∞/∞) for the first two terms, with the third term equaling zero. In order to figure out what the limit is as b goes to ∞, we can just apply L' Hopital's Rule on each of the terms. Applying L'Hopital's Rule: lim b ---> ∞: [ (-2x / 4(e^2x)) - (1 / 4(e^2x)) - 0 ] - [ -(1/4) ] Plugging in '∞' in for x still gives us the indeterminate form of (∞ / ∞) for the first term. We therefore need to apply L'Hopital's Rule once again: lim b ---> ∞: [ (-2 / 8(e^2x)) - 0 - 0 ] + (1/4) Plugging in '∞' for all values of x gives us: lim b ---> ∞: [ -2 / b ] + 1/4 Final Answer: 1/4
Is infinity minus infinity zero?
There is no single number that is infinity. For example, in mathematics, there are an infinite number of integers, but there are also an infinite number of real numbers. However, there are infinitely more real numbers than integers, but because there are infinite real numbers and integers, there must be different values for infinity. If you want a more proof-like explanation, then here’s a proof that infinity-infinity=undefined. Let’s start by assuming that infinity minus infinity is 0. In equation form, that is [math]infinity-infinity=0[/math]. Then, we can use one of my favorite infinity theorems: [math]infinity+infinity=infinity[/math]. Subtracting infinity from both sides in that equation gives you [math]infinity=infinity-infinity[/math]. Since we said earlier that [math]infinity-infinity=0[/math], we can substitute in zero for the RHS of that equation, giving us in[math]finity=0[/math], which is obviously not true. We have just proved that infinity-infinity does not equal zero, nor any other number. Infinity-infinity is simply undefined.
The limit of (1/n), as n goes to infinity is zero, but (1/n) is not equal to zero when n goes to infinity, where is the fallacy or the philosophy?
6.7.2016 — “The limit of [math]\frac{1}{n}[/math], as n approaches infinity is zero, but [math]\frac{1}{n}[/math] is not equal to zero when n approaches infinity, where is the fallacy or the philosophy?” — Thank you for asking me to answer this question.If a sequence has a limit, it is not necessary for the terms of the sequence to ever equal the limit. In fact this point is crucial to the definition of limit because many sequences do not attain their limit (if they have one) and because the idea of limit is not about a value at one point but in neighborhoods of a point. Informally, all that is needed for a limit to exist as ’n approaches infinity’ is that the terms of the sequence to get as close to zero as desired by choosing n large enough (most calculus texts will give you a formal definition).Thus in the sequence 0. 0, 0, … the limit is 0 and every term equals the limit. For the sequence of the question, no term is zero but the limit is zero.A good question is why limits are defined this way. It is because this definition allows a rigorous approach to estimation of limits of sequences, e.g. of partial sums and partial fractions (and limits allow rigorous treatment of functions in calculus). It also allows ‘arithmetic’ of sequences — adding, subtracting, multiplying, dividing, and more.
What is the integral of cos(x) from 0 to infinity?
The area under one arch of the cosine curve is exactly 2. If the arch is below the x-axis the area is -2. Only one half of the first arch is included, since it is bisected by the y-axis. So the total area must be: 1 - 2 + 2 - 2 + 2 - 2 +2 - 2 + 2 - .... = 1 + (-2+2) + (-2+2) + (-2+2) + ... = 1 + 0 + 0 + 0 + 0 + .... = 1. On the other hand, by the associative property of addition this same sum equals: (1-2) + (2-2) + (2-2) + (2-2) + .... = -1 + 0 + 0 + 0 + 0 + .... = -1. So the area must be 1. Or -1. Or maybe 1 = -1. Or maybe the associative property of addition does not work for infinite sums. Or maybe we have to develop a more careful theory of infinite series, or of integrals from 0 to infinity.
Is the integral from 0 to infinity of xe^-x dx = 1 ?
Integrating by parts, INT x*e^(-x) dx = -x*e^(-x) + INT e^(-x) dx = -x*e^(-x) - e^(-x) + c Now, plug in the bounds (0 to infinity) and then evaluate. lim(x->infinity) [-x*e^(-x) - e^(-x)] - [0 - e^(0)] = 0 - [-1] = 1 So, yes, the answer is 1.
Improper integral and hopitals rule:the integral from infinity to 1 of:5ln(x)/x^4dx?
Hi, Pizzadude. Your antiderivative is correct. But there is no reason to use L'Hopital's Rule. If the integral is from 1 to infinity, we can evaluate it as the limit as b goes to infinity as the integral from 1 to b of: ∫ 5 ln(x) / x^4 dx Integrating from 1 to b, we are left with: [ -5((3 ln(b) + 1) / 9b^3) ] - [ -5((3 ln(1) + 1) / 9(1)^3) ] Since the leftmost term goes to 0 as b goes to infinity, this is: = 0 - (-5 / 9) = 5 / 9 -- Note, if the integral really is from infinity to 1, and not the other way around, then, since the function is bound on [1, infinity), we may simply reverse the sign, and the solution is -5 / 9.
What is infinity over zero?
Thanks for the A2A!This makes absolutely no sense.Infinity isn’t a number, it’s a concept.Division by [math]0[/math] is undefinedClearly [math]\dfrac{\infty}{0}[/math] is nonsense. Even if you get it when trying to evaluate a limit, it’s indeterminate. Try doing the limit a different way.
What is infinity to the power zero?
What is Infinity to the power zero?[math]\infty^0[/math] is an indeterminate form, but[math]\lim\limits_{n\to\infty}n^0=\lim\limits_{n\to\infty}1=1[/math]In general [math]\lim\limits_{f(x)\to\infty,g(x)\to0}f(x)^{g(x)}[/math]is indeterminate as it depends on how rapidly the functions approach infinity and zero compared to each other.As stated, for the simple functions [math]f(n)=n, g(n)=0[/math] the result is a constant sequence as [math]n[/math] increases and so has a limit of the constant, namely [math]1[/math].
What is zero times infinity?
If you're thinking of infinity as a number, then zero times infinity is not defined. You'll often find it given as NaN, "not a number", on a computer.If you're thinking of it as a form of a limit, it's an indeterminate form, and the limit may or may not exist. As a limit, it appears as [math]\displaystyle\lim_{x\to a}\,f(x)g(x)[/math] where [math]\displaystyle\lim_{x\to a}\,f(x)=0[/math] and [math]\displaystyle\lim_{x\to a}g(x)=\infty[/math]. Here are three examples.[math]\displaystyle\lim_{x\to 0}x^2\csc x[/math]. As [math]x[/math] approaches [math]0[/math], the first term of the product, [math]x^2[/math] approaches [math]0[/math], while the second term, the cosecant of [math]x[/math], approaches [math]\pm\infty[/math]. As [math]x[/math] approaches [math]0[/math] from the positive side, [math]\csc x[/math] approaches [math]+\infty[/math], but as [math]x[/math] approaches [math]0[/math] from the negative side, [math]\csc x[/math] approaches [math]-\infty[/math]. The product, however, approaches [math]0[/math]. You can check that by taking values of [math]x[/math] near [math]0[/math], and asking your calculator to evaluate [math]x^2\csc x[/math]; you'll see that the product is almost the same as the [math]x[/math] you started with, so as [math]x\to 0[/math], [math]x^2\csc x\to0[/math] also.[math]\displaystyle\lim_{x\to 0}x^2\csc^4 x[/math]. Although this limit is of the same form, [math]0\cdot\infty[/math], it approaches [math]\infty[/math].[math]\displaystyle\lim_{x\to 0}x^2\csc^2 x[/math]. The value of this limit is 1.Those three examples show that different limits of the form [math]0\cdot\infty[/math] can be [math]0,\infty,[/math] or [math]1[/math]. Other limits of this form can have negative values, and others don't exist because the product oscillates or has some other strange behavior.Because you can't tell the value of limits of this form by only looking at the form, this form is called an indeterminate form. To determine their actual values you need to do more analysis.