# Thread: differentiation and intergration of sinx/x and d/dt P(t)=P(t)

1. ## differentiation and intergration of sinx/x and d/dt P(t)=P(t)

(1) problem

differentiation [intergration of sint/t], limit from 0 to x^2

let say f(t)=t, F(t)=t^2/2, if f(t)=sint/t, F(t)=intergration of sint/t

d/dx [intergration of sint/t] limit from 0 to x^2
d/dx [F(t)] limit from 0 to x^2
{d/dx [F(t)} = differentiation of F(t)

then {d/dx [F(t)} = [f(t)] limit from 0 to x^2

f(x^2)-f(0)
(sin x^2/x)-(sin0/0)

(sin x^2/x)-1

in this correct?

(2) problem

d/dt P(t)=P(t)

t=0, P(0)=0

t->infinite, then P(infinite)=f(h)/h

in this right?

2. Originally Posted by toumah12
(1) problem

differentiation [intergration of sint/t], limit from 0 to x^2

let say f(t)=t, F(t)=t^2/2, if f(t)=sint/t, F(t)=intergration of sint/t

d/dx [intergration of sint/t] limit from 0 to x^2
d/dx [F(t)] limit from 0 to x^2
{d/dx [F(t)} = differentiation of F(t)

then {d/dx [F(t)} = [f(t)] limit from 0 to x^2

f(x^2)-f(0)
(sin x^2/x)-(sin0/0)

(sin x^2/x)-1

in this correct?
Given that $\displaystyle F^{\prime}(t)=f(t)$, $\displaystyle \frac{\,d}{\,dx}\int_a^{u(x)}f(t)\,dt=\frac{\,d}{\ ,dx}\left[F(u(x))-F(a)\right]=F^{\prime}(u(x))\cdot u^{\prime}(x)-0=f(u(x))\cdot u^{\prime}(x)$

So it follows that $\displaystyle \frac{\,d}{\,dx}\int_0^{x^2}\frac{\sin t}{t}\,dt=\frac{\sin(x^2)}{x^2}\cdot 2x=\frac{2\sin(x^2)}{x}$

(2) problem

d/dt P(t)=P(t)

t=0, P(0)=0

t->infinite, then P(infinite)=f(h)/h

in this right?
I'm not quite sure what you're trying to do here...

3. thanks a lot.

(2) problem is a part of question.

d/dt P(t)=P(t)

t=0, P(0)=0

t->infinite, then P(infinite)=f(h)/h

in this right?
---End Quote---

I'm not quite sure what you're trying to do here...

(2) problem is only sub part of the full question that is to find the P(t) or Q(t) when t=0 and t->infinite

d/dt P(t) = -aP(t)+bQ(t)--1
d/dt Q(t) = aP(t)-bQ(t)--2
P(t)+Q(t) = 1

a and b are real number.

let d/dt P(t) = P(t), d/dt Q(t) = Q(t)

from the eq 1 and 2

Q(t)=[a-Q(t)]/[a+b]---3
P(t)=[b-P(t)]/[a+b]---4

from the eq 3 and 4

when t=0, P(t)=0 or Q(t)=0
Q(t)=[a-0]/[a+b]
P(t)=[b-0]/[a+b]

P(t)+Q(t) = 1, (a/[a+b])+(b/[a+b])=1 {prove}

thus when t=0, Q(t)=a/(a+b) or P(t)=b/(a+b)

but, i was wondering if t->infinite, what would be the value of P(t)?

Could this [f(h+a)-f(a)]/h to be used?

lim t->infinite [P(h+infinte)-P(infinite)]/h

P(h)/h in this correct?

recall eq 4
P(t)=[b-P(h)/h ]/[a+b] will this be answer?