# How to integrate

Discussion in 'Math' started by jag1972, Nov 4, 2011.

1. ### jag1972 Thread Starter Active Member

Feb 25, 2010
71
0
Hello Folks,
Could someone please confirm I am thinking along the right track here?

I have derived a function for determine the DC power stored in a capacitor for a series RC circuit.
The function looks like this: p (t) = VI*exp^mt(1-exp(mt)).
I would like to integrate this to obtain the energy stored in the capacitor. Would this have to be done using by substitution or by parts method?

Thank you in advance.

2. ### someonesdad Senior Member

Jul 7, 2009
1,585
142
Assuming you meant $p(t) = V I e^{mt}(1 - e^{mt})$, note this is just the sum of two exponentials. Thus, it's amenable to substitution (assuming you want to integrate with respect to t). You might want to read the introduction to using TeX in your posts -- it's pretty straightforward to learn and results in nicer-looking formulas.

3. ### jag1972 Thread Starter Active Member

Feb 25, 2010
71
0
Thanks someonesdad,
I am little confused about the whole process, I can integrate this function by substitution.
p(t)=VI*e$^{mx}$(1-e$^{mx}$)

let u = 1-e$^{mx}$ and $\frac{du}{dx}$=-me$^{mx}$

$\frac{du}{-m}$ = e$^{mx}$
Every part can now be substituted

$p(t) = \frac{-VI}{m}\int u du$

$\frac{-VI}{m}$ $\frac{ u^{2}}{2}$

$\frac{-VI}{m}$(1-2$^ e{mx}$+ $^e{mx}$)

Could I just not integrate this equation normally for example.

p(t)=VI*e$^{mx}$(1-e$^{mx}$) is equal to

p(t)=VI*e$^{mx}$- VI*e$^{2mx}$

Integrate both sides or is this not valid?

4. ### someonesdad Senior Member

Jul 7, 2009
1,585
142
That's exactly what I meant when I said it's the sum of two exponentials. Beginning math students sometimes don't understand the importance of $\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$, but you obviously do.

jag1972 likes this.
5. ### jag1972 Thread Starter Active Member

Feb 25, 2010
71
0
Thank you very much for your help

6. ### jag1972 Thread Starter Active Member

Feb 25, 2010
71
0
I have integrated the following function using integrals (integrating both sides and using u-substitution): $p(t) = VI (e^{mx}$ $- e^{2mx}$)

I am very sorry about this however I can not get both equations to match up perfectly. I would really appreciate some advice.

The sum of integral equation: $e(t) = \frac{VI e^{mx}}{m} - \frac{VI e^{2mx}}{2m} ^ {t2}_{t1}$

in a reduced form it is: $e(t) = \frac{VI e^{mx}}{2m} ( 2 - {e^{mx}} )^ {t2}_{t1}$

However when I use U - substitution the result is:

$e(t) = \frac{-VI e^{mx}}{2m} ( 1 - 2{e^{mx}} + {e^{2mx}})^ {t2}_{t1}$

I have tried and tried to get them to match but I can not, I have enetered both functions into MATLAB and they both give the same result. I attached the code and screen shot of output.

Thank you in advance.

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7. ### jag1972 Thread Starter Active Member

Feb 25, 2010
71
0
I wrote the last equation incorrectly it should have been:

$p(t) =\frac{-VI}{2m}(1 - 2e ^{mx}+ e^{2mx})$

Multiply both sides by -1

$p(t) =\frac{VI}{2m}(-1 + 2e ^{mx}- e^{2mx})$

When this integral is expanded the first term in the bracket -1 drops out leaving:

$p(t) =\frac{VI}{2m}(2e ^{mx}- e^{2mx})$

or

$p(t) = \frac{VIe^{mx}}{2m}(2 - e^{mx})$

This is in the same form as the first integration method. Whohoo!

I think I will have a cup of tea to celebrate

8. ### someonesdad Senior Member

Jul 7, 2009
1,585
142
Even better, you figured out what you did wrong yourself. This is so much more conducive to learning than just having someone point out a mistake. Congratulations! And I hope you enjoyed the tea.

jag1972 likes this.