Integration: Question 4

Worked solution

Part (a): Integrate dy/dx to find the equation of the curve

Rewrite the gradient function as powers of $x$:

$\frac{dy}{dx} = 12x^2 - 6x^{-2}$

Integrate term by term:

$\int 12x^2\,dx = 4x^3 \qquad \int -6x^{-2}\,dx = -6 \times \frac{x^{-1}}{-1} = 6x^{-1} = \frac{6}{x}$

So, including a constant of integration:

$y = 4x^3 + \frac{6}{x} + c$

Use the given point $(1, 9)$. Substitute $x=1$, $y=9$:

$9 = 4(1)^3 + \frac{6}{1} + c = 4 + 6 + c = 10 + c \implies c = -1$

So the equation of the curve is

$y = 4x^3 + \frac{6}{x} - 1$

Recompute independently by differentiating this result back:

$\frac{d}{dx}\left(4x^3 + \frac{6}{x} - 1\right) = 12x^2 + 6\times(-1)x^{-2} - 0 = 12x^2 - \frac{6}{x^2}$

This matches the given $\dfrac{dy}{dx}$ exactly. Also re-check the point: at $x=1$, $y = 4(1) + 6(1) - 1 = 4+6-1=9$ ✓, matching $(1,9)$. Both checks confirm $c=-1$ is correct.

Part (b): Evaluate y when x = 3

$y = 4(3)^3 + \frac{6}{3} - 1 = 4(27) + 2 - 1 = 108 + 2 - 1 = 109$

Recompute independently: $3^3 = 27$, and $4 \times 27 = 108$ (since $27 \times 4 = 27\times2\times2 = 54\times2=108$); $6\div3=2$; so $108+2-1=109$, matching the first calculation.

Final answers

• (a) $y = 4x^3 + \dfrac{6}{x} - 1$
• (b) $y = 109$ when $x=3$