Integral Calculator

A definite integral computes the exact signed area between a function f(x) and the x-axis over the interval [a, b]. When an exact antiderivative is known, the Fundamental Theorem of Calculus gives the exact answer, and this calculator works that way: it recognises a handful of standard forms, writes down the antiderivative, and evaluates it at both ends. Nothing here is approximated.

Fundamental Theorem of Calculus: ∫ₐᵇ f(x) dx = F(b) − F(a) where F’(x) = f(x)

Common antiderivative rules:

• ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C (power rule, n ≠ −1)
• ∫ eˣ dx = eˣ + C
• ∫ sin(x) dx = −cos(x) + C
• ∫ cos(x) dx = sin(x) + C
• ∫ 1/x dx = ln|x| + C

What each variable means:

• a, b: lower and upper bounds of integration
• f(x): the integrand (the function being integrated)
• F(x): the antiderivative of f(x)
• C: the constant of integration, which every indefinite integral carries

Worked example: ∫₁⁴ (3x² − 2x + 1) dx Antiderivative F(x) = x³ − x² + x F(4) = 64 − 16 + 4 = 52 F(1) = 1 − 1 + 1 = 1 Result = 52 − 1 = 51

Enter that one as a polynomial with a = 3, b = −2, c = 1 and bounds 1 to 4, and you should get 51 back.

Signed area, not area. A definite integral counts area below the x-axis as negative. ∫₀^2π sin(x) dx is zero, not 4, because the hump above the axis and the trough below it cancel exactly. If you want the geometric area regardless of sign, integrate |f(x)|, or split the interval at each root and add the absolute values.

Where the Fundamental Theorem stops working. It requires f to be continuous on the whole of [a, b]. Push that requirement and you get answers that look fine and are not. Take ∫₋₁¹ x⁻² dx. The antiderivative is −1/x, so F(1) − F(−1) = −1 − 1 = −2. But x⁻² is positive everywhere it is defined, so a negative area is impossible. The catch is the vertical asymptote at x = 0 sitting inside the interval: the integral genuinely diverges, and the −2 is an artifact of applying a theorem outside its conditions. This calculator checks for that case and says so rather than printing the number.

The same trap catches ∫₋₁¹ (1/x) dx, where the two halves cancel to a tidy 0. That is a Cauchy principal value, not a convergent integral, and treating it as one is a well-worn way to derive nonsense.

The +C is not decoration. An indefinite integral is a whole family of functions, since any constant differentiates to zero. It cancels out in a definite integral, which is why the constant never appears in the bounded answer, but leaving it off an indefinite one is a genuine error rather than a formatting choice.