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find m(x) if m(0) = 8 and m'(x) = x^2 * e^(x^3)
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a u sub should get it put \(u=x^3, du=3x^2dx\) and you are almost done
@satellite73 help me with my question next please?
you get \[\frac{1}{3}\int e^udu = \frac{1}{3}e^u+C=\frac{1}{3}e^{x^3}+C\] then solve for \(C\) since \[m(0)=\frac{1}{3}+C=8\]
@brittneyy18rm i'll look at it
Thanks a bunch!
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