integration help
- Thread starter synx
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- Feb 4, 2004
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Your formatting is somewhat confusing. Are you inquiring about the potential equality of the following two expressions?
. . . . .\(\displaystyle \large{\int_1^0 \.\left(\frac{d}{dx}\, e^{\arctan{(x)}}\,\right)\,dx}\)
. . . . .\(\displaystyle \large{\frac{d}{dx}\, \int_1^0\, e^{\arctan{(x)}} \,dx}\)
When you reply, please include all of the work and reasoning you have tried thus far. Thank you.
Eliz.
. . . . .\(\displaystyle \large{\int_1^0 \.\left(\frac{d}{dx}\, e^{\arctan{(x)}}\,\right)\,dx}\)
. . . . .\(\displaystyle \large{\frac{d}{dx}\, \int_1^0\, e^{\arctan{(x)}} \,dx}\)
When you reply, please include all of the work and reasoning you have tried thus far. Thank you.
Eliz.
Yep, those questions, except the 0 on bottom and 1 on top of integral.
For the second one I believe the d/dx and integral cancel eachother out giving just arctan(x).
For the first, I'm not really sure if i'm doing it right, but I just did d/dx e^arctan(x) which is e^arctan(x)/x^2+1, then integrate that which brings back e^arctan(x) and do e^arctan(1) - e^arctan(0), which equals e^(pi/4) -1 .
For the second one I believe the d/dx and integral cancel eachother out giving just arctan(x).
For the first, I'm not really sure if i'm doing it right, but I just did d/dx e^arctan(x) which is e^arctan(x)/x^2+1, then integrate that which brings back e^arctan(x) and do e^arctan(1) - e^arctan(0), which equals e^(pi/4) -1 .
