IP AMaths Notes (Upper Sec, Year 3-4): 17) Integration Essentials

Integration is the inverse of differentiation. Start with antiderivatives and practice evaluating definite integrals.

Fundamental rules

• Reverse power: \( \int x^n , dx = \tfrac{1}{n + 1} x^{n+1} + C \) for \( n \neq -1 \).
• Log form: \( \int \tfrac{1}{x} dx = \ln |x| + C \).
• Constant multiple and sum integrate term by term.

Worked example 1 — Definite integral

Evaluate \( \int_1^3 (4x^{2} - 6x + 5) , dx \).

1. Antiderivative: \( \tfrac{4}{3} x^{3} - 3x^{2} + 5x \).
2. Substitute bounds: \( \Bigl[\tfrac{4}{3} x^{3} - 3x^{2} + 5x\Bigr]_1^3 \).
3. Compute: at \( x = 3 \), value is \( \tfrac{4}{3} \times 27 - 27 + 15 = 36 - 27 + 15 = 24 \).
4. At \( x = 1 \), value is \( \tfrac{4}{3} - 3 + 5 = \tfrac{4}{3} + 2 \).
5. Difference: \( 24 - \bigl(\tfrac{4}{3} + 2\bigr) = \tfrac{72}{3} - \tfrac{10}{3} = \tfrac{62}{3} \).

Worked example 2 — Substitution

Evaluate \( \int_0^1 6x (x^{2} + 1)^2 , dx \).

1. Let \( u = x^{2} + 1 \); then \( \dfrac{du}{dx} = 2x \) so \( x , dx = \tfrac{1}{2} du \).
2. When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = 2 \).
3. Integral becomes \( \int_1^2 6 \times \tfrac{1}{2} u^{2} , du = 3 \int_1^2 u^{2} , du \).
4. Evaluate: \( 3 \Bigl[ \tfrac{1}{3} u^{3} \Bigr]_1^2 = \Bigl[ u^{3} \Bigr]_1^2 = 8 - 1 = 7 \).

Try this

Integrate \( \int (5x^4 - 3x + 2) , dx \) and state the constant of integration explicitly.