43. Find the equation of the curve that flas a gradient of \( (3 x-1)^{3} \) and passes through point \( \left(-\frac{1}{3}, \frac{4}{3}\right) \).

To find the equation of the curve, we start by integrating the gradient function \( (3x - 1)^3 \). This gives us: \[ y = \int (3x - 1)^{3} \, dx \] Using substitution and integration techniques, we can find the antiderivative, and then we'll determine the constant by using the point \( \left(-\frac{1}{3}, \frac{4}{3}\right) \). Carrying out the integration gives: \[ y = \frac{(3x - 1)^4}{12} + C \] To find \( C \), we'll substitute \( x = -\frac{1}{3} \) into our equation and set \( y = \frac{4}{3} \): \[ \frac{4}{3} = \frac{(3(-\frac{1}{3}) - 1)^4}{12} + C \] Solving this will lead us to the constant \( C \). This process will yield the specific expression for \( y \), giving us the required equation of the curve. Now let’s have some fun with additional knowledge! The history of calculus is rich and fascinating! It was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century. They created the foundations for derivatives and integrals, which are pivotal in understanding the change and accumulation of quantities—essentially giving us the tools to study everything from physics to economics. If you find integration tricky, you're not alone! A common mistake is neglecting substitution when it's necessary or not properly handling the constants of integration. Always remember to check your work: differentiate your result to ensure it matches the original gradient, and don’t skip the step of plugging in points to find constants! This will help reinforce your understanding and prevent errors.