Additional Mathematics 0606 / Calculus — Integration / 1.541.54: Finding a curve from second derivative0606/12/O/N/19 — Question 11 · 8 marksMark as done · Save for later · Show all solutionsA curve is such that d2ydx2=2(3x−1)−23\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2(3x - 1)^{-\frac{2}{3}}dx2d2y=2(3x−1)−32. Given that the curve has a gradient of 666 at the point (3,11)(3, 11)(3,11), find the equation of the curve.[8]▸ Answer▸ Official mark scheme← 1.23: Equation of a curve from a trigonometric second derivative1.82: Product rule and reverse integration →