Sauberschrauber

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1 years, 61 days

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These are answers submitted by Sauberschrauber

restart; fn1 := int((a-x).lambda.exp(-`λx`), x = 0 .. a)

int(`.`(a-x, lambda, exp(-`λx`)), x = 0 .. a)

(1)

fn2 := int((x-a).lambda.exp(-`λx`), x = a .. infinity)

int(`.`(x-a, lambda, exp(-`λx`)), x = a .. infinity)

(2)

f2 := sum((b1*fn1+b2.fn2)/lambda, lambda = 1 .. c)

sum((b1*(int(`.`(a-x, lambda, exp(-`λx`)), x = 0 .. a))+b2.(int(`.`(x-a, lambda, exp(-`λx`)), x = a .. infinity)))/lambda, lambda = 1 .. c)

(3)

temp1 := sum((b1*(int(`.`(a-x, lambda, exp(-`λx`)), x = 0 .. a))+b2.(int(`.`(x-a, lambda, exp(-`λx`)), x = a .. infinity)))/lambda, lambda = 1 .. c); 'temp1'

temp1

(4)

diff(temp1, a)

sum((b1*(lambda.exp(-`λx`))*a-infinity*(b2.signum(lambda.exp(-`λx`))))/lambda, lambda = 1 .. c)

(5)

solve({sum((b1*(lambda.exp(-`λx`))*a-infinity*(b2.signum(lambda.exp(-`λx`))))/lambda, lambda = 1 .. c)}, [a])

[[a = infinity*(sum((b2.signum(lambda.exp(-`λx`)))/lambda, lambda = 1 .. c))/(b1*(sum((lambda.exp(-`λx`))/lambda, lambda = 1 .. c)))]]

(6)

NULL

Download integration_doubt_exp.mw

Dear vicky2811,

I tried the file and got an error message about e and exp() .

Did you mean exp () ? Then the result looks different ..

I am not sure if that is what you want ?

Best regards, 

Klaus

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