Calculus Integral Question

Please note

$\displaystyle A = \frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2 = \frac{(b - a^2 c)^3} {6 a^3 c^2}$

Taking derivatative wrt. $c$ and equating to zero,

$\displaystyle A' = 0 = \frac{2(b-a^2c)^3 + 3ca^2(b-a^2c)^2}{6a^3c^3}$

This gives us $\displaystyle c = \frac{b}{a^2}, c = - \frac{2b}{a^2}$

Tags:

Integration

Calculus

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