Asked by Anonymous
Calculate the following definite integrals:
(a)∫x cos(x) dx
(b)∫x ln(x) dx
(c)∫xe−x^2dx
Notice that this last integral can be computed by substitution.
What happens if you use parts?
(a)∫x cos(x) dx
(b)∫x ln(x) dx
(c)∫xe−x^2dx
Notice that this last integral can be computed by substitution.
What happens if you use parts?
Answers
Answered by
mathhelper
You clearly want to use "integration by parts" for a) and b)
Among the best of math youtubes you can find are those by
"blackpenredpen". He shows a beautiful little algorithm that cuts through
all the fog for the steps needed to integrate by parts. He calls it the DI method. Here is the link, (make sure you delete the spaces at the front of the link, I inserted 4 spaces in the link)
ht tp s://ww w.youtube.c om/watch?v=2I-_SV8cwsw&ab_channel=blackpenredpen
make sure you look at all 3 different outcomes.
I came across this method from one of my profs over 60 years ago at the University of Waterloo in Canada, but have not found it in any of the current Calculus books.
Enjoy, your question become very easy with this.
Among the best of math youtubes you can find are those by
"blackpenredpen". He shows a beautiful little algorithm that cuts through
all the fog for the steps needed to integrate by parts. He calls it the DI method. Here is the link, (make sure you delete the spaces at the front of the link, I inserted 4 spaces in the link)
ht tp s://ww w.youtube.c om/watch?v=2I-_SV8cwsw&ab_channel=blackpenredpen
make sure you look at all 3 different outcomes.
I came across this method from one of my profs over 60 years ago at the University of Waterloo in Canada, but have not found it in any of the current Calculus books.
Enjoy, your question become very easy with this.
Answered by
oobleck
∫xe(−x^2) dx
Let u = e^(-x^2) du = -2x e^(-2x^2) dx
dv = x dx, v = 1/2 x^2
∫xe(−x^2) dx = 1/2 x^2 e^(-x^2) + ∫ x^3 e^(-x^2) dx
You can see that this is going nowhere fast
Let u = e^(-x^2) du = -2x e^(-2x^2) dx
dv = x dx, v = 1/2 x^2
∫xe(−x^2) dx = 1/2 x^2 e^(-x^2) + ∫ x^3 e^(-x^2) dx
You can see that this is going nowhere fast
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