• Rewrite roots & reciprocals as powers before integrating: \(\sqrt{x} = x^{1/2},\ \frac{1}{x^3}=x^{-3}\).
• Never integrate \(x^{-1} = \frac{1}{x}\) in Year 12.
• No +C for area or volume.
• For area below the x-axis: make the answer positive.
• For volume: square the function before integrating.

Example A — Equation of a Curve

Given: \(\frac{dy}{dx} = 4x^{-2} - 3\), and the curve passes through (1, 5).

\[ y = \int (4x^{-2} - 3)\, dx = -4x^{-1} - 3x + C \] Substitute \(x=1, y=5\): \[ 5 = -4 - 3 + C \Rightarrow C = 12 \] Final: \(y = -4x^{-1} - 3x + 12\).

Example B — Area Between Curve and x-axis

Find the area under \(y = x^2 - 4x\) between intercepts.

\(x^2 - 4x = x(x-4) = 0 \Rightarrow x=0,4.\) \[ \text{Area} = \int_0^4 (4x - x^2) dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = \frac{32}{3} \] Area = \( \frac{32}{3} \) units².

Example C — Volume by Rotation

Find the volume when \(y = \sqrt{x}\) is rotated from \(x=1\) to \(x=4\).

Rewrite: \(\sqrt{x} = x^{1/2} \Rightarrow y^2 = x\) \[ V = \pi \int_{1}^{4} x\, dx = \pi \left[\frac{x^2}{2}\right]_{1}^{4} = \pi \cdot \frac{15}{2} \] Volume = \( \frac{15\pi}{2} \) units³.

• Always rewrite before integrating.
• Check whether +C is needed.
• State limits clearly in definite integrals.
• Square the function first for volume, never afterwards.
• Draw a quick sketch if unsure about area signs.