Finding the Equation of a Curve from Its Gradient

To recover a curve from its gradient function, you reverse differentiation, integrate, and then use a known point to fix the constant of integration. The $+C$ is the whole point of the question: without it you have a family of parallel curves, not the one you want.

A curve passes through the point $(1, 5)$ and has gradient function $\dfrac{dy}{dx} = 6x^2 - 4x + 3$. Find the equation of the curve. [5]

The working

Step 1, integrate $\frac{dy}{dx}$ to get $y$ (remember the $+C$): $y = \int (6x^2 - 4x + 3)\,dx = 2x^3 - 2x^2 + 3x + C \quad \text{(M1, A1)}$

Step 2, use the point $(1, 5)$ to find $C$, substitute $x = 1$, $y = 5$: $5 = 2(1)^3 - 2(1)^2 + 3(1) + C = 2 - 2 + 3 + C = 3 + C \quad \text{(M1)}$ $\Rightarrow C = 2 \quad \text{(A1)}$

Step 3, write the equation: $y = 2x^3 - 2x^2 + 3x + 2 \quad \text{(A1)}$

Where the marks are won and lost

• Integrate correctly: raise each power by one and divide, $6x^2 \to 2x^3$, $-4x \to -2x^2$, $3 \to 3x$. A slip here derails everything.
• The $+C$ is essential, and it’s an accuracy mark. Omitting it means you can’t use the point and the answer is incomplete.
• Use the point to solve for $C$; don’t leave the answer as "$\dots + C$". The question asks for the curve, which needs a numerical $C$.

Common mistakes

• Differentiating instead of integrating (going the wrong way).
• Forgetting the constant of integration.
• Finding $C$ but not writing the final equation with it substituted in.

Full method: Integration notes. Topic home: Calculus pillar.