techniques of integration

label Calculus
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the voltage v (in volts) induced in a tape head is given by v=t^2e^3t, where t is the time (in seconds). find the average value of v over the interval from t=0 to t=2. Round to the nearest volt.

i need the work!!

Jun 12th, 2015

Thank you for the opportunity to help you with your question!

Given the voltage v=t^2 e^3t

to find the average value of v in the interval from t=0 to t=2 is

average value of v = integral  v  dt

                               = integral ( t=0 to t=2) t^2 e^3t  dt

                               = ( t=0 to t=2) (1 / 3)t^2*e^(3t) - (2 / 9)t*e^(3t) + (2 / 27)e^(3t)

     now substitute value of t


we get = [(1/3)(2)^2 * e^6 - (2/9)2  * e^(3 *2) + (2/27)e^(3*2) ] -[ (1/3)(0)^2 *e^(3*0) -(2/9)0*e^(3*0) +(2/27)e^(3*0)

          = (1/3)4*e^6 - (2/9)2e^6 + (2/27) e^6  -  [0 -0+(2/27)*1 ]

          =  4/3 * e^6 -4/9 e^6 +2/27e^6 -2/27

         = [ 4*9 e^6 - 4*3 e^6 +2 e^6 -2 ] / 27

        = [ 36 e^6 - 12e^6 + 2 e^6 - 2 ] / 27

       = [ 46 e^6 - 2]/27

so average value is [ 46 e^6 - 2]/27

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 12th, 2015

Hi user , I'm new to this study pool. If  you have any doubts , Just comment

Jun 12th, 2015

it tells me incorrect!!

Jun 12th, 2015

one sec i will convert the answer into  numerical value, Then it will work

Jun 12th, 2015

one sec i will convert the answer into  numerical value, Then it will work

Jun 12th, 2015

ok thanks... i need the steps, thanks again!

Jun 12th, 2015
ok the steps of integration right??
Jun 12th, 2015

yes...

Jun 12th, 2015

so ok first i will expalain how to get the answer

average value of v = [46 e^6 -2 ]/27

                               =[ 46*404.961 - 2 ] / 27

                               =689.86 volts

 rounding average value will be 690 volt

Jun 12th, 2015

it saids incorrect!

Jun 12th, 2015
ok then i will do it again
Jun 12th, 2015

ok thanks

Jun 12th, 2015

integration part


Jun 12th, 2015
img_20150612_210306060.jpg

so once you are done with the integartion

then  we have to plug t=2 and t =0

so

 =(1/3) t^2 * e^3t  - (2/9) t * e^3t + (2/27)e^3t  this what we got, now plug t values

= [ (1/3) (2)^2 * e^(3*2)  -  (2/9) 2* e^(3*2) + (2/27) e^(3*2) ] -  [ 0 -0 + 2/27 ]

= [ 4/3 e^6 - 4/9 e^6 + 2/27 e^6 - 2/27]

= [4*9 e^6 - 4*3 e^6 +2 e^6 -2 ]/27

=[36 e^6 - 12 e^6 + 2 e^6 - 2 ] /27

= [ 26 e^6 - 2 ] / 27

so average value will be  = [ 26 e^6 - 2 ] / 27*2

                                        = [ 26 e^6 - 2]/ 54

Jun 12th, 2015

it would be approximately  195 volt

Jun 12th, 2015

194 will be fine?

Jun 12th, 2015

if you have any doubt plug e =2.72 and calculate like 194.something

Jun 12th, 2015
do you understand the procedure?


Jun 12th, 2015
do you understand the procedure?


Jun 12th, 2015

yes!! thank you very much!!

Jun 13th, 2015

thanks a lot . If you have any questions in maths just ping me .

Jun 13th, 2015

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