The gradient of a curve is given by $\dfrac {dy}{dx}=3x^{2}+a$, where a is a constant. The curve passes through the points (−1, 2) and (2, 17). Find the equation of the curve.

The gradient of the curve is $\dfrac {dy}{dx}=3x^{2}+a$

$\therefore$ the equation of the curve would be given by $\therefore \int \left( 3x^{2}+a\right) dx$

$\therefore y=\int \left( 3x^{2}+a\right) dx$

$\therefore y=3\times \dfrac {1}{3}\times x^{3}+ax+k$

$\therefore y=x^{3}+ax+k$

The curve passes through the points (−1, 2) and (2, 17).

At (-1,2): $2=\left( -1\right) ^{3}+a\left( -1\right) +k$

or, $2=-1-a+k$, $\therefore a=k-3$

At (2,17): $17=2^{3}+a\times 2+k$, or $17=8+2a+k$

$\therefore 2a=9-k$

$\therefore 2(k-3)=9-k$ or $2k-6=9-k$

$\therefore 3k=15$ or $k=5$

As $a=k-3$ or $a=5-3$ or $a=2$

As $x^{3}+ax+k$

$\therefore$ the equation of the curve is $x^{3}+2x+5$