Integral Calculator

How the Integral Calculator Works

The integral calculator computes both definite and indefinite integrals using fundamental rules of calculus. An indefinite integral (antiderivative) finds the family of functions whose derivative is the given function: ∫f(x)dx = F(x) + C. A definite integral calculates the net area under a curve between two bounds: ∫[a to b] f(x)dx = F(b) − F(a), using the Fundamental Theorem of Calculus.

Example: ∫x²dx = x³/3 + C (indefinite). Evaluated from 1 to 3: [3³/3] − [1³/3] = 9 − 0.333 = 8.667. This represents the exact area under the parabola y = x² between x = 1 and x = 3.

Integration Rules: The Core Antiderivative Formulas

Essential integration rules every calculus student needs:

• Power rule: ∫xⁿdx = xⁿ⁺¹/(n+1) + C (for n != −1)
• Constant rule: ∫k dx = kx + C
• Natural logarithm: ∫(1/x)dx = ln|x| + C
• Exponential: ∫eˣdx = eˣ + C; ∫aˣdx = aˣ/ln(a) + C
• Trigonometric: ∫sin(x)dx = −cos(x) + C; ∫cos(x)dx = sin(x) + C
• Inverse trig: ∫dx/√(1−x²) = arcsin(x) + C; ∫dx/(1+x²) = arctan(x) + C
• Sum/difference rule: ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx
• Constant multiple: ∫cf(x)dx = c∫f(x)dx

Definite Integral Calculator: Finding Area Under a Curve

The definite integral ∫[a to b] f(x)dx represents the net signed area between the function and the x-axis from x = a to x = b. "Net signed" means areas above the x-axis are positive; areas below are negative and can cancel out.

Worked examples with common functions:

• ∫[0 to 2] 3x²dx = [x³] from 0 to 2 = 8 − 0 = 8
• ∫[0 to π] sin(x)dx = [−cos(x)] from 0 to π = (−cos π) − (−cos 0) = 1 + 1 = 2
• ∫[1 to e] (1/x)dx = [ln x] from 1 to e = ln(e) − ln(1) = 1 − 0 = 1
• ∫[0 to 1] eˣdx = [eˣ] from 0 to 1 = e − 1 ~ 1.718

When the function dips below the x-axis within the integration bounds, the definite integral gives net area (positive minus negative regions). To find total (unsigned) area, split the integral at zero-crossings and take the absolute value of each part.

Advanced Integration Techniques

When basic rules don't directly apply, these techniques handle more complex integrals:

• U-substitution: For composite functions. Let u = inner function, substitute and simplify. Example: ∫2x·(x²+1)⁵dx → let u = x²+1, du = 2x dx → ∫u⁵du = u⁶/6 + C = (x²+1)⁶/6 + C
• Integration by parts: For products of functions. ∫u dv = uv − ∫v du. LIATE rule for choosing u: Logarithms > Inverse trig > Algebraic > Trig > Exponential. Example: ∫x·eˣdx → u = x, dv = eˣdx → x·eˣ − ∫eˣdx = xeˣ − eˣ + C
• Partial fractions: For rational functions (polynomial/polynomial). Decompose into simpler fractions before integrating.
• Trigonometric substitution: For integrals involving √(a²−x²), √(a²+x²), or √(x²−a²). Substitute x = a·sin(θ), x = a·tan(θ), or x = a·sec(θ) respectively.

Frequently Asked Questions

What is the difference between definite and indefinite integrals?

An indefinite integral produces a function (plus constant C): ∫x²dx = x³/3 + C. A definite integral produces a number by evaluating the antiderivative at two specific bounds: ∫[1 to 3] x²dx = [x³/3] from 1 to 3 = 9 − 1/3 = 26/3 ~ 8.667. The indefinite integral is the general antiderivative; the definite integral uses the Fundamental Theorem to evaluate exact area. The +C cancels in definite integrals: F(b) + C − (F(a) + C) = F(b) − F(a).

What does the antiderivative represent?

The antiderivative F(x) of f(x) is the function whose derivative is f(x). In physics: if f(x) is velocity, F(x) is position. If f(x) is acceleration, F(x) is velocity. In economics: if f(x) is marginal cost, F(x) is total cost. The antiderivative "undoes" differentiation and is the cornerstone of the Fundamental Theorem of Calculus, which connects the apparently separate concepts of derivatives and areas.

Why do I add +C to indefinite integrals?

The derivative of any constant is zero. So both x³/3 and x³/3 + 7 have derivative x². Without +C, you'd only find one specific antiderivative rather than the complete family. In definite integrals, this doesn't matter because the constant cancels. In differential equations and physics problems, the +C is determined by initial conditions — the specific value of C that satisfies the boundary conditions of the problem.