/sci/ - Science & Math

[Anonymous No.10940569]

I don't understand equation (1.21), especially the [math]e^{-x^3}[\math] part, could someone explain it?
Also it's cool if you can point me to some resource that explain how to differentiate when the [math]x[\math] is in the [math]\int[\math], like
[eqn]\frac{d}{dx} \int _0 ^x f(y) dy[\eqn]

[Anonymous No.10940606]

>>10940569
Say you have a function of the kind [math]g(x) = \int_0^x f(y) dy [\math]. Suppose f(x) has a primitive (antiderivative) F(x), that is [math] \int f(x) dx = F(x) + C [\math] in the sense of indefinite integration. Then [math] g(x) = \int_0^x f(y) dy = F(x) - F(0) [\math]. The derivative of g(x) w.r.t x is the derivative of F(x), which is f(x).

[Anonymous No.10940613]

>>10940569
>>10940606
Why are you both using "[\math]" instead of "[/math]" ?

[Anonymous No.10940755]

>>10940613
>>10940606
Cause I'm a retard.

Say you have a function of the kind [math]g(x) = \int_0^x f(y) dy [/math]. Suppose f(x) has a primitive (antiderivative) F(x), that is [math] \int f(x) dx = F(x) + C [\math] in the sense of indefinite integration. Then [math] g(x) = \int_0^x f(y) dy = F(x) - F(0) [/math]. The derivative of g(x) w.r.t x is the derivative of F(x), which is f(x).

[Anonymous No.10941461]

>>10940569
Use 1.20 in picture:
(1.1) dF(t,t)/dt = dF(t',t'')/dt'+dF(t',t'')/dt''
The differential of dF(t,t) is likewise
dF(t,t) = dF(t',a)+dF(b,t'') or expressed in terms of (1.1)
dF(t,t) = f(t',t'')dt'+f(t',t'')dt''
Integrate both sides to obtain
F(t,t) = Int (f(t',t'')dt' + Int f(t',t'')dt'' where Int stands for Integration.
Make the integrals dependent on x rather than on t. We get Int(f(x,x)) = Int(f(t',t'')dt' + Int(f(t',t'')dt''
Consider case (1.21).
Since the result of Int(dte^-xt^2) only depends on x, we may equate differentiation by dx with dt. Now f(t',t') here is dt' and e^-xt''^2. We already know that Intf(t,t)dt = Int(t',a)dt'+Int(b,t'')dt''. a and b are assumed t according to (1.20). They are actually not.
We get d(Int(dt*e^-xt^2))/dx = d(Int(dt'))e^-xa^2/dt'+d(Int(e^-xt''^2)db). We get d(Int(dt*e^-xt^2)/dx = e^-x^3-Int(dt*t^2*e^-xt^2) as expected.

[Anonymous No.10941655]

>>10940755
hah u did id aggain

[Anonymous No.10943427]

>>10940569
do you know what the derivative [math]\frac{d}{dx} \int_a^x f(t)\, dt[/math] is ?

[Anonymous No.10943710]

>>10940569
This is from my book. Did you buy this book or not? I can request your IP and sue you for piracy. You have to reply now.

[Anonymous No.10943748]

>>10943710
Uhhh whatch out, we got a big boy here. Kys larping faggot.

[Anonymous No.10945490]

>>10943710
the book is free for download here http://www.physics.miami.edu/~nearing/mathmethods/
stupid fuck

[Anonymous No.10945668]

OP here

>>10941461
Your explanation is obscure
But anyway I found a much simpler approach with this https://en.wikipedia.org/wiki/Leibniz_integral_rule