Solution to that weird u-substitution, blackballwhitechalk

filmed location: UC Berkeley Evans Hall (i.e. math building)

Solution to that weird u-substitution,
math for fun,
calculus 1 integral example,
calculus 2 integral example,
u-substitution, integral of sqrt(1+x^2),

[wellbangok]

I'll just stick with the trig sub lmao.

[zhupdhk]

One may also show _sqrt(x^2+1)_ is an even function, therefore the area bounded under its graph on the interval [-1, 1] is twice the area bounded on the interval [0, 1].

[gooball]

4:32 I have a negative
I have one half
uh
negative one half

[weerman]

Now you'll have to change your channel name to 'whitechalk'

[sergpodolnii]

I would use the function symmetry properties for redefining the boundaries.

[mokouf]

I was wondering why you did this u-sub in the first place, but I watched to the end.
Finally, "This u-sub sucks!" is what I agree. A wrong u-sub just wastes your time!

[jaredbeaufait]

Just do x=tanu then it becomes integral of sec^3 which is just integration by parts

[arekkrolak]

good video! I like how you explain your viewers how to claculate integrals and in the same video you explain the same viewers than half and half equals one ;)

[ralfbodemann]

I would use symmetry considerations.
The integrand f(x)=(1+x^2)^1/2 is an even function.
The integration interval is symmetrical around 0.
Therefore, the integral of f(x) from -1 to 1 equals 2 time the integral of f(x) from 0 to 1 (or from -1 to 0).

[marcelweber]

Neither black chalk nor red chalk, but at least the 8-ball mic makes me sleep happily :)

[ahmedamin]

sir, you were intgrating an even function from -1 to 1 is the same 2*intgral from 0 to 1

[vivekjha]

We can simplify the prob by recognizing that f(x) = (1 + x^2)^.5 is an even function {i.e. f(-x) = f(x)}. Then we can simplify by:

integral(f, -1 to 1) = 2 * integral(f, 0 to 1)

This way we can "disregard" the negative square root value of x and achieve the same answer more elegantly.

[holyshit]

Integrals of the binomial differentials
Tschebyshov substitution can be used in integral you have got

We have integral

\int{u^{1/2}(-1+u)^{-1/2}du}
m=1/2
n=1
p=-1/2
U{m+1}{n}+p=U{3}{2}-U{1}{2}=1 in Z
so we can use substitution v^2=\frac{u-1}{u}

[mrKreuzfeld]

one way of explaining the split is; The function F(X) = U must be invertible and on (ie, for every value U there is one and only one X, and for every value X there is one and only one U). The function U = X^2 + 1 does not fulfill that requirement. That is why you have to split the positive and the negative functions.

[ishtiaqhussain]

Your Vidz with blackboard are much more satisfying though I appreciate all your vidz.

[DjVortex-w]

So while trying to solve the integral, you just made it more complicated.

[laurierbaribeau]

I'm not convinced the 0 is nonsense. It's not really what you want, but 0 seems like a valid answer considering the square root can put out + and -. The question could be from 0 to 1 to guarantee the interesting part. I think the math was accepting the negative version of the square root graph for x < 0 and the positive version for x >0, so the "area under the curve" could sum to zero. The u sub wasn't lying.... it just wasn't helping.

[Infinitesap]

Im beginning to understand this :)
Your material is awesome.
Could you elaborate more in depth about the changes in domain (I guess). I think the coupling to the chain rule is what I still don't see clearly when I glance at a integral. Is it because the integrals bare made up so that the always have a solution. Sort of a puzzle where all the pieces has been designed beforehand and thus the puzzle can be solved?

[Kalimaco]

Hello could you teach how to deduce the metric of the half-plane of Poincaré

[technicallightingfriend]

Can you do a challenging questions of integral, dx/(sinx-1)^2.