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About LinkBacksOriginally Posted by alexandrabel90
}e^x \sin\left(e^x\right)=\int_1^\pi \sin\left(z\right)dz=-\cos(\pi)+\cos(1)=1+\cos(1))
My first reaction was to wonder why you think it should be! But I think I see. I suspect you made the substitution u= e^{x} so that the integral becamedu= \left(cos(u)\right)_0^{ln(\pi)})
.
Well, there are two errors in that. The first, small error, is that is the integral from 0 to, not the other way around. I suspect you didn't mean it that way, anyway.
The second, more important, error is that the original limits of integration, ln(x) and 0, apply to the original variable, x, not u! There are two ways of handling that correctly.
The first is to ignore the limits of integration until the end. As an indefinite integral,. Now, change back to the variable x. Since
, that is the same as
. At the upper limit,
. At the lower limit, 0, that is
. Subtracting,
which is what Drexel28 got.
The other, better way, to do this is to change the limits of integration as you go. When x= 0,. When
,
. Now we have that
[tex]= \left(-cos(u)\right)_{u=1}^\pi= (-cos(\pi))- (-cos(1))= 1+ cos(1) again.
That's one reason I advocate writing the limits of integration in the form "x= ..." as I did in
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