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Mathematics 8 Online
OpenStudy (anonymous):

Integral of x=y^2?

OpenStudy (anonymous):

(1/3)y^3

OpenStudy (anonymous):

with respect to x *

OpenStudy (anonymous):

oh so then is the question integral sqrt(x) dx ?

OpenStudy (unklerhaukus):

\[x=y^2\] \[\int x\text dx=\int y^2\text dx\] \[\frac {x^2}2=y^2x+c\]

OpenStudy (unklerhaukus):

hmmm i think have made a false assumption \(y=y(x)\)

OpenStudy (unklerhaukus):

\[y^2=x\]\[y=\pm \sqrt x\] \[\int y\text dx=\int\pm\sqrt x\text dx\]\[xy=\pm\int x^{1/2}\text dx\]\[xy=\pm \frac{x^{3/2}}{3/2}+c\]\[xy=\frac 23x^{3/2}+c\]\[y=\frac 23 x^{1/2}+c\]\[y=\frac 23 \sqrt{x}+c\]

OpenStudy (unklerhaukus):

ops \[xy=\pm\frac 23 x^{3/2}+c\]\[y=\pm\frac 23\sqrt x+\frac cx\]

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