Question

The slope at any point of a curve y = ƒ(x) is given by $\frac{dy}{dx} = 3x^{2}$ and it passes through (-,1). The equation of the curve is

• A. $y = x^{3} + 2$
• B. $y = - x^{3} + 4$
• C. $y = 3x^{3} + 4$
• D. $y = - x^{3} - 2$

Answer 1

Answer: (A)

$y = x^{3} + 2$

Answer 2

$\frac{dy}{dx} = 3x^{2} \Rightarrow \int_{}^{}{dy} = \int_{}^{}{3x^{2}dx} \Rightarrow y = x^{3} + c$

$\therefore$ The curve passes through (-1,1)

$\therefore$= - 1 + c $\Rightarrow$ c = 2 $\Rightarrow$ y = x³ + 2.