Weird u-substitution for the integral of sqrt(1+x^2)

Integral by u sub,
integral of sqrt(1+x^2),
integral of sqrt(1+x^2) from -1 to 1,

blackpenredpen

[ehtuanK]

Wow, that makes solving integrals so much easier! When taking the intergral over [a, b] just use a u-sub with u=x(x-a-b) and you get an integral over [-ab, -ab] which is always zero! Thx for the pro-tip :)

On a more serious note: I wonder how many integrals I already messed up by applying the u-sub the wrong way.

[blackpenredpen]

Keep in mind, this video is about this u-sub step. I know how to integrate sqrt(1+x^2) with.... 3 different ways

[blurginouliz]

The problem lies in the fact that he didn't write down the integrand after the substitution... when you do u=1+x^2 and du=2xdx, when you solve for dx: dx=du/(2x), you have to eliminate that x of course, so you go back to u=1+x^2 and solve for x, and when you do that, you get x=+_sqrt(u-1), and so you broke the integrand into 2 separate functions now, each with different domains, so since x goes from -1 to 1, you have to do it from -1 to 0 for one function, and then from 0 to 1 for the other function. In the u domain it will go from 2 to 1, then from 1 to 2. i'll write it down: (integral of [ -(1/2)sqrt(u/(u-1))]du from 2 to 1)+(integral of [(1/2)sqrt(u/(u-1))]du from 1 to 2). if you invert the domain of integration on the first integral and change the sign of the integrand, then adding both the integrals together, you get the final integral: integral of [sqrt(u/(u-1))]du from 1 to 2. And that is not equal to 0.

[ObitoSigma]

The u-sub is an improper integral because the range is not defined on the range (-1, 1) in the u-world.

u=1+x^2
x=sqrt(u-1)
du=2xdx
dx=du/(2sqrt(u-1))

.5*∫u/sqrt(u-1)du

[BigDBrian]

can't you prove any integral is equal to 0 with this method? for instance you could use u=2+(x-a)(x-b), a and b being the ends of the range of the integral, which would always make it an integral from 2 to 2 and therefore be 0.

[terdragontra]

Given that this proof would imply EVERY integral from -1 to 1 is zero (and furthermore, with some fiddling, every possible integral is zero), i can safely say the u-sub step is wrong.

[technicbrasil]

Well that must be false, since the square root will always give us positive values there is no way to cancel out the `positive area` with a `negative area` below the x-axis, however I am not quite sure why the u-sub cannot be applied on that situation

[johnfeild]

Excellent example to demonstrate the need to be careful of the domain.

The integrand is even, so int(-1, 1, sqrt(1+x^2) dx) = 2*int(0, 1, sqrt(1+x^2) dx). Although everything is fine in the x world, we find that the integrand is not defined at the lower limit in the u world.

[lakshaymd]

well we know it can't be 0 because √(1+x²) is always positive so it should be something positive :/

[erdemekinci]

The trick you used is not working as when you make substitution u=f(x), f must be injective over the interval [a, b]. Actually, you can make any integral zero may making the wrong substitution. Set U=(x-a)(x-b) then the integral from a to b turns to be integral from zero to zero which is zero!!!

[samuelbam]

the function sqrt(1+x^2) is symmetric at the y-axis because of the x^2. When you do the integral form -1 to 1 you have to add the integral from 0 to 1 and the integral from -1 to 0, but the integral form -1 to 0 is just negative the integral from 0 to 1 so 0 is the right answer

[OblivionCalling]

The function is symmetric about the y-axis so the integral from -1 to 0 is equal to the integral from 0 to 1. We can use this to rewrite the question

integral [-1, 1] f(x) dx = integral [-1, 0] f(x) dx + integral [0, 1] f(x) dx

integral [-1, 1] f(x) dx = 2 * integral [0, 1] f(x) dx

u = x^2 + 1
u = (0)^2 + 1 = 1
u = (1)^2 + 1 = 1

2 * integral [0, 1] f(x) dx = 2 * integral [1, 2] f(u) du

[saschatrumper]

int(sqrt(1+x²), -1, 1) = int(sqrt(1+x²), -1, 0) + int(sqrt(1+x²), 0, 1) = int(sqrt(u), 2, 1) + int(sqrt(u), 1, 2)
because 1+x² is an even function [ f(-x) = f(x) ] but u is an odd function [ f(-u) = - f(u) ].

You cant substitute an even function with an odd function, as long your interval crosses the axis of symmetry.

[MrRoyalChicken]

The squareroot of x will always give you a positive value while in this case what you need is the negative value. When squaring a number and then taking the squareroot the information about whether the number was positive or negative is lost, since the x^2 function is not a one to one mapping from the real numbers to the real numbers but from the real numbers to all the positive real numbers.

[linkvssonic]

The function inside the integral must be injective in the closed interval [a, b] where the integral is applying. As we can see in the video: sqrt(x^2+1) is not injective in such interval. Therefore, we get the false answer 0.

[modestorosado]

Clearly there's a problem, because the integrand is a strictly positive function, so there is no way the integral could be zero. The issue is that 1+x^2 is not an injective function, which is why the substitution doesn't work and messes up the whole integral.

[TheShingwa]

Actually, could it be because it is an invalid sub? If you integrating between the limits 1 and -1, from fiddling with the differentials, you get 1/2(sqrt(u-1))du=dx. Since the limits change to u=2 you get a negative under the square root.

[holyshit]

Function to substitute should be one to one to be able to change interval
Integral will be zero if integrand is odd function

[Capnarchie]

yes, it has to be 0 because it's a defined integral using F(b)-F(a) and since b and a are 2 it will be 0 after you get to the subtraction part

[siddartharcot]

if you 'x' make it in terms of 'u' you get x= plus minus 2root(u-1), in order to get x=1 you have to use x=2root(u-1) and x=-1 would be x= -2root(u-1), as you cant integrate a plus minus the integral would be invalid from the start