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4 answers

Let u = e^x -1, then du = e^x dx, so you have the following in terms of u:

⌡u^-.5 du = 2 u^.5
= 2 sqrt (e^x - 1) from 0 to ln 3
= 2 [sqrt (e^ln3 - 1) - sqrt (e^0 - 1)
= 2sqrt 2 - 2 sqrt 0 = 2 sqrt 2

2007-03-29 04:56:10 · answer #1 · answered by Kathleen K 7 · 0 0

2*sqrt(2)

hint:
put [ (e^x) -1] = t

2007-03-28 19:37:51 · answer #2 · answered by qwert 5 · 0 0

then why no longer me ? The greater the merrier ! .. ? x d(?x) .................... on [ 0, one million ] = ? (?x)² d(?x) ............... on [ 0, one million ] = [ (?x)³ / 3 ] ................. on [ 0, one million ] = (one million/3) [ (?one million)³ - (?0)³ ] = (one million/3) [ one million - 0 ] = one million/3 ..................................... Ans. .........................................

2017-01-05 10:13:52 · answer #3 · answered by bolander 3 · 0 0

for(i=0 to ln3)
{
.........................;
Y={e to the power x divided by(square root ...)};
.........................;
}

2007-03-28 17:10:03 · answer #4 · answered by Muhammad+Islam+25 1 · 0 1

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