Integration & Definite Integral Calculator

Integration Guide

The Power Rule for Integration

∫ axⁿ dx = ax^(n+1)/(n+1) + C (for n ≠ −1). Examples: ∫ 3x² dx = x³ + C. ∫ 5 dx = 5x + C. ∫ x^(1/2) dx = (2/3)x^(3/2) + C. ∫ 1/x² dx = ∫ x^(−2) dx = −x^(−1) + C = −1/x + C. Exception: ∫ 1/x dx = ln|x| + C (the n=−1 case). Definite integral: [F(x)] from a to b = F(b) − F(a). Area from 0 to 3 of 3x²: [x³] from 0 to 3 = 27 − 0 = 27.

Standard Integrals

∫ sin(ax) dx = −(1/a)cos(ax) + C. ∫ cos(ax) dx = (1/a)sin(ax) + C. ∫ e^(ax) dx = (1/a)e^(ax) + C. ∫ 1/x dx = ln|x| + C. ∫ sec²(ax) dx = (1/a)tan(ax) + C. The mnemonic: integration reverses differentiation — each standard integral can be verified by differentiating the result. A−level: these results should be memorised. They are given in examination formula sheets at some awarding bodies (check yours).

Areas Under Curves

Definite integral = signed area (negative below x-axis). For areas: if f(x) < 0 over [a,b], the integral is negative but the area is positive. To find total unsigned area: identify where f(x) = 0 (x-intercepts). Integrate each region separately. Add absolute values. Example: y = x³ − x from x=0 to x=2. y crosses zero at x=0 and x=1. ∫ from 0 to 1: negative region = area = 1/4. ∫ from 1 to 2: positive = [x⁴/4 − x²/2] = 4−2−(1/4−1/2) = 2.25. Total area = 0.25 + 2.25 = 2.5 square units.

Trapezium Rule (Numerical Integration)

When a function cannot be integrated analytically: area ≈ (h/2)[y₀ + 2(y₁+y₂+...+y_{n-1}) + y_n]. Where h = strip width = (b−a)/n. n strips gives n+1 ordinates. More strips → more accurate. For y = √(1+x³) from 0 to 2, n=4: h = 0.5. Ordinates at x = 0, 0.5, 1, 1.5, 2: y = 1, 1.061, 1.414, 1.837, 3.000. Area ≈ (0.5/2)[1 + 2(1.061+1.414+1.837) + 3] = 0.25[1 + 8.624 + 3] = 0.25 × 12.624 = 3.156. This is the standard GCSE/A-level numerical method — required when integration by substitution is impossible.