Integration is the reverse of differentiation. For Year 12, we only integrate functions that can be written as \(kx^n\), including roots and reciprocals (except \(x^{-1}\)).
If \(n \neq -1\), then:
We cannot use this rule when \(n = -1\). That case is \( \dfrac{1}{x} \) and is not included in Year 12.
Example 1: \(\displaystyle \int x^4\,dx\)
\[ \int x^4\,dx = \frac{1}{5}x^5 + C \]
Example 2: \(\displaystyle \int 3x^{-2}\,dx\)
\[ 3 \int x^{-2} dx = 3\cdot\frac{x^{-1}}{-1} + C = -3x^{-1} + C = -\frac{3}{x} + C \]
Example 3 (Root Function): \(\displaystyle \int \sqrt{x}\,dx\)
Rewrite first: \( \sqrt{x} = x^{1/2} \) \[ \int x^{1/2} dx = \frac{x^{3/2}}{\frac{3}{2}} + C = \frac{2}{3}x^{3/2} + C \]
Example 4 (Multiple Terms): \(\displaystyle \int \left(4x^3 + 5x^{-2} - \sqrt{x}\right) dx\)
Rewrite: \( \sqrt{x} = x^{1/2} \) \[ \int \left(4x^3 + 5x^{-2} - x^{1/2}\right) dx = x^4 - 5x^{-1} - \frac{2}{3}x^{3/2} + C \]
If we know \( \frac{dy}{dx} \), we integrate to find \(y\).
Example: \( \dfrac{dy}{dx} = 6x^2, \ y = 8 \text{ when } x = 1\).
\[ y = \int 6x^2 \, dx = 2x^3 + C \] Substitute \(x=1, y=8\): \[ 8 = 2(1)^3 + C \Rightarrow C = 6 \] Final equation: \(y = 2x^3 + 6\)
Area between curve \(y=f(x)\) and the x-axis from \(a\) to \(b\):
Example: Find area under \(y= x^2\) from \(x=0\) to \(x=2\).
\[ \int_0^2 x^2 dx = \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{8}{3} \]
Volume when a curve is rotated about the x-axis:
Example: Find the volume generated by rotating \(y=x\) from \(x=0\) to \(x=1\).
\[ V = \pi \int_0^1 x^2 dx = \pi \left[\frac{x^3}{3}\right]_0^1 = \frac{\pi}{3} \]