Question

Integration: How would I get to this answer?

Answer 1

Given that $y$ attains a maximum at $x=1$, it follows that $y'=0$ at that same point. So integrating once gives

$\displaystyle\int\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\int-8x\,\mathrm dx$
$\dfrac{\mathrm dy}{\mathrm dx}=-4x^2+C_1$
$\implies -4(1)^2+C_1=0\implies C_1=4$

and so the first derivative is $y'=-4x^2+4$.

Integrating again, you get

$\displaystyle\int\frac{\mathrm dy}{\mathrm dx}\,\mathrm dx=\int(-4x^2+4)\,\mathrm dx$
$y=-\dfrac43x^3+4x+C_2$

You know that this curve passes through the point (2, -1), which means when $x=2$, you have $y=-1$:

$-1=-\dfrac43(2)^3+4(2)+C_2$
$\implies C_2=\dfrac53$

and so

$y=-\dfrac43x^3+4x+\dfrac53$