202211041123 Solution to 2004-CE-AMATH-II-3

The slope at any point $(x,y)$ of a curve $C$ is given by

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2+1}$ .

If the $x$ -intercept of $C$ is $1$ , find the equation of $C$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (3x^2+1)\,\mathrm{d}x \\ y(x) & = x^3 + x + C\textrm{ for some constant} \\ \end{aligned}$

Substituting $(1,0)$ for $(x,y(x))$ :

$\begin{aligned} 0 & = (1)^3+(1) + C \\ C & = -2 \\ \end{aligned}$

In conclusion, the equation of $C$ is

$y(x) = x^3 + x -2$ .

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