Posted by **tiff** on Thursday, August 27, 2009 at 5:29am.

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!

- Math -
**Writeacher**, Thursday, August 27, 2009 at 8:10am
assistance needed

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- integration -
**drwls**, Thursday, August 27, 2009 at 8:39am
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.

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