Integration & Definite Integral Calculator

Integration Guide

What should I know about 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.

What's the key thing to understand about 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).

What do I need to know about 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.

What should I know about 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, 2.092, 3.000. Area ≈ (0.5/2)[1 + 2(1.061+1.414+2.092) + 3] = 0.25[1 + 9.134 + 3] = 0.25 × 13.134 = 3.28. This is the standard GCSE/A-level numerical method — required when integration by substitution is impossible.

What do I need to know about Integration by Substitution?

When an integral contains a function and something close to its derivative, substitution simplifies it to a standard form. For ∫2x·(x²+1)⁵ dx, let u = x²+1, so du = 2x dx — which conveniently matches the 2x already present. The integral becomes ∫u⁵ du = u⁶/6 + C = (x²+1)⁶/6 + C. The method works by recognising the chain rule running in reverse: if the integrand looks like f'(x)·g(f(x)), substitution peels away the outer function. Common triggers are an expression raised to a power multiplied by its own derivative, or e^(something) multiplied by the derivative of that something. This is the counterpart to the power rule shown elsewhere on this page — the power rule handles simple x^n terms directly, while substitution extends it to composite functions, and it's the first technique taught after basic integration in A-Level Maths and most first-year calculus courses.

What should I know about Definite vs Indefinite Integrals?

An indefinite integral ∫f(x)dx returns a family of functions plus an arbitrary constant C, because any constant disappears when you differentiate — ∫3x²dx = x³+C, and C could be anything. A definite integral ∫[a to b]f(x)dx instead evaluates to a single number: compute the antiderivative F(x), then calculate F(b)−F(a); the constant C cancels out entirely, which is why it's dropped for definite integrals. For ∫[0 to 2]3x²dx: F(x)=x³, so F(2)−F(0) = 8−0 = 8. This number represents the signed area between the curve and the x-axis across that interval — area below the axis counts as negative, which is why a definite integral of a function that dips negative can give a smaller (or negative) total even though the curve clearly bounds visible area. Separating 'find the antiderivative' from 'evaluate at the limits' is the cleanest way to avoid sign errors in definite integral problems.

What should I know about Integration by Parts?

Integration by parts is used when a function to integrate is a product of two simpler functions that can't be combined with the power rule or basic substitution alone. It's derived from the product rule for derivatives and states ∫u dv = uv - ∫v du, where you choose one factor to call u (ideally one that simplifies when differentiated) and the rest to call dv (ideally one that's easy to integrate). A classic example is ∫x·eˣ dx: letting u = x (so du = dx) and dv = eˣdx (so v = eˣ) gives x·eˣ - ∫eˣ dx = x·eˣ - eˣ + C. The "LIATE" rule of thumb — preferring Logarithmic, Inverse trig, Algebraic, Trigonometric, then Exponential functions as u, in that order — helps decide which factor to choose when it isn't obvious.

Worked Example: Area Under a Curve Using Definite Integration

To find the exact area between a curve and the x-axis over an interval, integrate the function to get its antiderivative, then evaluate that antiderivative at the upper and lower limits and subtract. Example 1 — a polynomial: find the area under y = 3x² + 2x between x = 1 and x = 4. The antiderivative is x³ + x² (adding 1 to each power and dividing by the new power: 3x² → x³, 2x → x²). Evaluate at the limits: at x = 4, that's 4³ + 4² = 64 + 16 = 80; at x = 1, it's 1³ + 1² = 1 + 1 = 2. The definite integral is 80 − 2 = 78 square units. Example 2 — a trig function: find the area under y = sin(x) between x = 0 and x = π. The antiderivative of sin(x) is −cos(x). At x = π: −cos(π) = −(−1) = 1. At x = 0: −cos(0) = −1. So the definite integral is 1 − (−1) = 2 square units — the familiar result that one full hump of a sine wave has an area of exactly 2.

Choosing the Right Integration Technique

Most integrals you meet can be tackled by matching the expression to one of a handful of standard techniques, roughly in this order of preference:

When an integral looks unfamiliar, it's usually worth trying substitution first (look for a 'chain rule in reverse' pattern), then integration by parts if it's clearly a product of two different function types, before resorting to a numerical method.

Improper Integrals

An integral is called improper when at least one limit of integration is infinite, or when the integrand becomes undefined (blows up) somewhere within the interval. These are evaluated as limits: for an infinite upper bound, the integral from a to infinity of f(x) dx is defined as the limit, as b approaches infinity, of the integral from a to b. For example, the integral from 1 to infinity of 1/x^2 dx equals the limit as b→∞ of [-1/x] from 1 to b, which is lim(b→∞) (-1/b + 1) = 0 + 1 = 1. This converges to a finite value even though the region extends forever, because the function shrinks fast enough. By contrast, the integral from 1 to infinity of 1/x dx diverges: [ln(x)] from 1 to b gives ln(b) - 0, which grows without bound as b→∞. This 1/x boundary case is the classic dividing line in calculus courses between integrals that converge and ones that do not, and it underlies tests like the integral test for infinite series convergence.

Common Integration Mistakes and How to Catch Them

A handful of errors account for most lost marks on integration problems, and each has a quick self-check.

The single most reliable check for any integration result is differentiation: if d/dx of your answer reproduces the original integrand exactly, the integral is correct, regardless of which technique was used to find it.