I'm asked to make a substitution and then evaluate the resulting integral in terms of area. I've tried letting u = 1-x^4 and letting u = x^2 both to no avail. I'm lost.
Any help would be appreciated!
Any help would be appreciated!
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Your second substitution which was u = x^2 seems correct.
∫x√(1 - x^4) dx from 0 to 1
u = x^2
du/2 = x dx
1/2*∫√(1 - u^2) from 0 to 1
Now note that the integral represents the area of a quarter circle of radius 1. Therefore, the answer is:
(1/2)(pi/4) =pi/8
∫x√(1 - x^4) dx from 0 to 1
u = x^2
du/2 = x dx
1/2*∫√(1 - u^2) from 0 to 1
Now note that the integral represents the area of a quarter circle of radius 1. Therefore, the answer is:
(1/2)(pi/4) =pi/8
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This question relies on knowing or looking up the standard integral
∫1/√(1 - u^2) du = sin^-1(u)
You correctly guessed substitution u = x^2
A common mistake here is to just replace x^2 by u but neglect the dx part
du/dx = 2x so 2x dx = du That helps a lot because we can rewrite
∫x/√(1 - x^4) dx = (1/2)∫1/√(1 - x^4)*2x dx = (1/2)∫1/√(1 - u^2) du
= (1/2)sin^-1(u) = (1/2)sin^-1(x^2)
Regards - Ian
∫1/√(1 - u^2) du = sin^-1(u)
You correctly guessed substitution u = x^2
A common mistake here is to just replace x^2 by u but neglect the dx part
du/dx = 2x so 2x dx = du That helps a lot because we can rewrite
∫x/√(1 - x^4) dx = (1/2)∫1/√(1 - x^4)*2x dx = (1/2)∫1/√(1 - u^2) du
= (1/2)sin^-1(u) = (1/2)sin^-1(x^2)
Regards - Ian
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suppose e try u=x^2
then we differentiate it and get du=2xdx
or
du/2=xdx
then your integral is of
0.5*sqrt(1-u^2)du
note: integration limits remain same
then we differentiate it and get du=2xdx
or
du/2=xdx
then your integral is of
0.5*sqrt(1-u^2)du
note: integration limits remain same