Finding a curve from its gradientEdexcel International A Level Maths: Revision notes

If you know the gradient function $\frac{\mathrm{d}y}{\mathrm{d}x}=f'(x)$ you can find the curve by integrating: $y=\int f'(x)\,\mathrm{d}x.$ The result contains an arbitrary constant $c$, so the gradient function alone describes a whole family of curves, each a vertical translation of the others. Example: $\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2-4x+1$ gives $y=x^3-2x^2+x+c$.

A point $(x_1,y_1)$ on the curve fixes $c$. Substitute the coordinates into the integrated expression and solve for $c$. Then write out the full equation $y=f(x)$. Example: the curve passes through $(2,9)$. Then $2^3-2(2^2)+2+c=9$, so $2+c=9$ and $c=7$. The curve is $y=x^3-2x^2+x+7$. Check: differentiating gives back $3x^2-4x+1$, and $x=2$ gives $y=9$.

The same method applies with roots and reciprocals. Rewrite as powers of $x$, integrate by adding one to the power and dividing, then use the point. Example: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{8}{x^3}-3\sqrt{x}=8x^{-3}-3x^{\frac12}$ gives $y=-4x^{-2}-2x^{\frac32}+c$. Through $(1,4)$: $-4-2+c=4$, so $c=10$ and $y=10-\frac{4}{x^2}-2x^{\frac32}$. At $x=4$, $x^{\frac32}=8$, so $y=10-\frac14-16=-\frac{25}{4}$.

If the gradient function is a product, expand it before integrating. For $f'(x)=(2x+3)(x-1)=2x^2+x-3$ we get $f(x)=\frac23x^3+\frac12x^2-3x+c$. Through $(3,10)$: $18+4.5-9+c=10$, so $c=-\frac72$. Two curves with the same gradient function differ only in $c$. For $\frac{\mathrm{d}y}{\mathrm{d}x}=6\sqrt{x}-\frac3{x^2}+2$ the general solution is $y=4x^{\frac32}+3x^{-1}+2x+c$. Through $(1,10)$, $c=1$. Through $(4,40)$, $c=-\frac34$. The curves are always $\frac74$ apart vertically.