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. 

Answer:

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 

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