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sqrt2 times e^t times t^(1/2) dt bounded between 0 and 1.
i tried integration by parts.. but it keeps repeating.. please help. thank you!
2 answers
Let's leave out the constant sqrt2 and do the rest. The constant can be multiplied back later.
I will use S to represent the integral sign.
Let u = e^t ; du = e^t dt
Let v = t^(1/2) ; dv = (1/2) t^(-1/2)
S u dv = e^t*t^(1/2) - S t^(1/2) e^t dt
= e^t*t^(1/2) - S u dv
Therefore
2 S u dv = e^t*t^(1/2)
S u dv = (1/2) e^t*t^(1/2)
Multiply that by sqrt2 for the integral you were asked for.
I will use S to represent the integral sign.
Let u = e^t ; du = e^t dt
Let v = t^(1/2) ; dv = (1/2) t^(-1/2)
S u dv = e^t*t^(1/2) - S t^(1/2) e^t dt
= e^t*t^(1/2) - S u dv
Therefore
2 S u dv = e^t*t^(1/2)
S u dv = (1/2) e^t*t^(1/2)
Multiply that by sqrt2 for the integral you were asked for.
0 answers