Finding a curve from dy/dx = 4x³ + 3/x⁴

Full worked solution

1. Integrate the gradient function to find the general form of $y$

   The gradient function is the derivative of $y$ with respect to $x$, so we integrate term-by-term to recover $y$, including a constant of integration $c$. Recall the integration rule: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne -1$.

   Rewrite the gradient function to use explicit powers for integration:

   $$\frac{dy}{dx} = 4x^3 + 3x^{-4}$$

   Integrate each term:

   $$\begin{align*} y &= \int \left(4x^3 + 3x^{-4}\right) dx \ &= 4 \cdot \frac{x^{4}}{4} + 3 \cdot \frac{x^{-3}}{-3} + c \ &= x^4 - x^{-3} + c \end{align*}$$

   This is the general form of the curve, valid for non-zero $x$.

   Note.

   When integrating negative powers, remember to add 1 to the exponent before dividing by the new exponent.

2. Use the given point to find the constant of integration $c$

   We know the curve passes through $(1,2)$, so substitute $x=1$ and $y=2$ into the general expression to solve for $c$:

   $$2 = (1)^4 - (1)^{-3} + c$$

   Simplify the terms: $1^4=1$ and $1^{-3}=1$, so:

   $$2 = 1 - 1 + c \implies c = 2$$

   Note.

   Always substitute the given point directly into the integrated form, not the original gradient function.

3. Rewrite the equation without negative indices

   Substitute $c=2$ back into the expression for $y$, and rewrite $x^{-3}$ as $\frac{1}{x^3}$ to eliminate negative indices:

   $$y = x^4 - \frac{1}{x^3} + 2$$

   This is the exact equation of the curve, with all terms in non-negative index form.

   Note.

   Double-check that all terms are valid for non-zero $x$, as specified in the problem.

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