50 points. Help please!
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Answered by
10
again using integration by substitution we can solve this ...
here
put limit to 1 to 4
2(e^√4 - e^√1)
= 2(e^2 -e )
= 2(e^2-e) ans
here
put limit to 1 to 4
2(e^√4 - e^√1)
= 2(e^2 -e )
= 2(e^2-e) ans
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Anonymous:
i have not put limits sorry
Answered by
29
we know ,
integration { f(x)ⁿ.df(x)/dx }dx = f(x)^(n+1)/(n+1)
use this concept here
let √x = z
differentiate
1/2√x dx = dz
dx = 2√x.dz
so, e^√x/√x dx convert into
2e^z.dz
we also know , integration of e^z = e^z
so, I =2 [ e^ ] put z = √x
I = 2[ e^√x ] put limit x = 1 to 4
I = 2 [ e^√4 - e^√1 ]
I = 2( e² -e)
hence, answer is 2(e² -e)
integration { f(x)ⁿ.df(x)/dx }dx = f(x)^(n+1)/(n+1)
use this concept here
let √x = z
differentiate
1/2√x dx = dz
dx = 2√x.dz
so, e^√x/√x dx convert into
2e^z.dz
we also know , integration of e^z = e^z
so, I =2 [ e^ ] put z = √x
I = 2[ e^√x ] put limit x = 1 to 4
I = 2 [ e^√4 - e^√1 ]
I = 2( e² -e)
hence, answer is 2(e² -e)
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