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### Network # integral derivative

#### kbcheong

Mar 24, 2012
25
if vC(t) = 1/C integral i(t)dt

with upper limit = t, lower limit = -infinity

if C = 0.001, i(t) = sin (1000*2PI*t)

I obtained the answer as

vC(t) = t*1000*sin(2000*Pi*t)

but the answer is seems incorrect, is it any website on integral calculus calculation?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,608
The question in itselft doesn't make sense from the start.
The indefinite integral: int(sin(a*t)dt) has the general solution -1/a*cos(a*t)
So the definite integral int(sin(a*t)dt from -infinity to t is
-1/a(cos((a*t)-cos(a*-infinity))
The latter term cos(a*-infinity) is undefined, so is the complete result. You need a definite value for the lower limit of the integral. e.g. lower limit=0 will give meaningful results.

A website is here: http://www.elainetron.com/apcalc/topic4.htm although you should know that much calculus from your classes.

Harald

#### kbcheong

Mar 24, 2012
25
I obtained the answer as -(cos(2000*t*PI))/(2*PI), may I enquire how to get the integration constant?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,608
1) your solution is nor completely right. You miss a factor of 1000 in the denominator. Compare with my post: what is "a"?
2) The integration constant comes from the lower integration limit, which here (-infinity) doesn't make sense.

Harald

#### Laplace

Apr 4, 2010
1,252
The difficulty here stems from the fact of dealing with an improper integral, in this case integrating over an unbounded region where the basic function never converges. Of course there are techniques for evaluating improper integrals but they all require convergence. Nevertheless, integrating from t=-infinity is theoretically correct but not applicable to a real physical system, i.e. the capacitor. If instead of minus infinity you begin the integration no further back than the creation of the fabric of space-time some 13.7 billion years ago, then evaluation of the definite integral will be possible.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,608
Possible: yes
But with a twist: depending on where exactly you start integrating the result at time t (today) will be different.This is because the indefinite integral is a periodic function.

Harald

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