CALTRX

Integral Calculator

Find antiderivatives of polynomial, trig, and exponential functions. See the complete solution with step-by-step working and formula explanations.

n
#
#

Enter your values above to see the results.

Tips & Notes

  • ✓Power rule: add 1 to the exponent, then divide by the new exponent. ∫x⁴ dx = x⁵/5 + C. ∫x^(-2) dx = x^(-1)/(-1) + C = -1/x + C. Works for all exponents except n = -1.
  • ✓Always add +C for indefinite integrals. The constant represents any vertical shift of the antiderivative family. Omitting it means your answer represents only one specific function, not the full solution.
  • ✓Definite integral: find the antiderivative F(x), then compute F(b) - F(a). The constant C cancels out. Example: ∫₀² x² dx = [x³/3]₀² = 8/3 - 0 = 8/3.
  • ✓U-substitution: identify a function and its derivative inside the integrand. Let u = inner function, find du. ∫2x·cos(x²) dx: let u = x², du = 2x dx, giving ∫cos(u) du = sin(u) + C = sin(x²) + C.
  • ✓Integration by parts: ∫u dv = uv - ∫v du. Choose u as the function that simplifies when differentiated (LIATE order: Logarithm, Inverse trig, Algebraic, Trig, Exponential). Example: ∫x·eˣ dx: u=x, dv=eˣdx, result = xeˣ - eˣ + C.
  • ✓Always verify by differentiating your answer. The derivative of the antiderivative must equal the original integrand. This is the most reliable way to check integration results.

Common Mistakes

  • ✗Forgetting +C for indefinite integrals. Every antiderivative is a family of functions differing by a constant. Without +C, the answer is technically incomplete and will cost marks in any exam.
  • ✗Power rule error: ∫x³ dx = x⁴/4 + C, not x⁴ + C. You must divide by the new exponent, not just increment it.
  • ✗Applying the power rule to ∫(1/x) dx. This integral is ln|x| + C, not x⁰/0 (which is undefined). The power rule fails specifically when the exponent becomes 0 after adding 1.
  • ✗Subtracting in the wrong direction for definite integrals: it is always F(b) - F(a), upper minus lower. Reversing this gives the correct magnitude but wrong sign.
  • ✗Missing the absolute value in ∫(1/x) dx = ln|x| + C. Without the absolute value, the result is undefined for negative x, making the antiderivative incomplete over its full domain.
  • ✗Forgetting to back-substitute after u-substitution in indefinite integrals. The final answer must be expressed in terms of the original variable x, not the substitution variable u.

Integral Calculator Overview

Integration is the reverse of differentiation and computes accumulated quantities — area under curves, total displacement from velocity, total charge from current, and net work done by variable forces. There are two forms: the indefinite integral (finding an antiderivative) and the definite integral (computing a specific numerical value). The Fundamental Theorem of Calculus, one of the most important results in all mathematics, reveals that these two seemingly different operations are inverses of each other.

Indefinite Integral — the antiderivative:

∫f(x)dx = F(x) + C, where F'(x) = f(x) and C is any constant
Power Rule for integration:
∫xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1
EX: ∫x³ dx = x⁴/4 + C | ∫x dx = x²/2 + C | ∫5 dx = 5x + C | ∫x⁻² dx = −x⁻¹ + C = −1/x + C
Definite Integral — area between curve and x-axis from a to b:
∫ₐᵇ f(x)dx = F(b) − F(a) (the Fundamental Theorem of Calculus)
EX: ∫₀³ x² dx = [x³/3]₀³ = 3³/3 − 0³/3 = 9 − 0 = 9 square units
Integration is fundamentally the reverse of differentiation, but the mechanics are more challenging — derivatives follow systematic rules that always produce an answer, while integrals of seemingly simple functions sometimes have no closed form at all. The standard antiderivative rules cover polynomials, exponentials, logarithms, and trigonometric functions; beyond these, substitution and integration by parts extend coverage to composed and product functions. Definite integrals compute net signed area — regions below the x-axis contribute negatively. When an integral spans regions both above and below the axis, the definite integral may be smaller than the actual total area. To find total area between a function and the x-axis, split the integral at each zero crossing and sum the absolute values of each piece. Numerical methods (Simpson's rule, Gaussian quadrature) handle integrals with no closed form, at the cost of approximate rather than exact results.

Frequently Asked Questions

Integration finds the antiderivative F(x) such that F′(x) = f(x), plus a constant C. Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1). ∫x³ dx = x⁴/4 + C. ∫x⁻¹ dx = ln|x| + C (the special case n=−1). ∫eˣ dx = eˣ + C. ∫sin(x) dx = −cos(x) + C. ∫cos(x) dx = sin(x) + C. The constant C represents any constant, since d/dx[F(x)+C] = f(x) for all C. Definite integrals eliminate C by evaluating F(b)−F(a) at the bounds.

The Fundamental Theorem of Calculus connects derivatives and integrals. Part 1: d/dx[∫ₐˣ f(t)dt] = f(x). Part 2: ∫ₐᵇ f(x)dx = F(b)−F(a) where F′=f. Example: ∫₁⁴ 2x dx. F(x) = x². F(4)−F(1) = 16−1 = 15. Verify geometrically: the region under y=2x from 1 to 4 is a trapezoid with area = (1/2)(2+8)(3) = 15 ✓. This theorem makes it possible to compute areas and accumulated quantities using antiderivatives rather than infinite sums.

U-substitution reverses the chain rule. Identify an inner function u=g(x), compute du=g′(x)dx, substitute to simplify, integrate in u, substitute back. Example: ∫2x·cos(x²)dx. Let u=x², du=2x dx. Integral becomes ∫cos(u)du = sin(u)+C = sin(x²)+C. Verify by differentiating: d/dx[sin(x²)] = cos(x²)·2x ✓. Integration by parts reverses the product rule: ∫u·dv = uv − ∫v·du. Choose u to be the term that simplifies when differentiated (LIATE order: Logarithm, Inverse trig, Algebraic, Trig, Exponential).

A definite integral ∫ₐᵇ f(x)dx represents the signed area between the curve y=f(x) and the x-axis from x=a to x=b. Signed means: area above the x-axis is positive, area below is negative. Example: ∫₀²π sin(x)dx = [−cos(x)]₀²π = −cos(2π)+cos(0) = −1+1 = 0. The positive area (0 to π) and negative area (π to 2π) cancel. For total area (ignoring sign), split at zeros: ∫₀π sin(x)dx + |∫π²π sin(x)dx| = 2+2 = 4.

Improper integrals have infinite limits or unbounded integrands. ∫₁∞ 1/x² dx = lim(b→∞) ∫₁ᵇ x⁻² dx = lim(b→∞) [−1/x]₁ᵇ = lim(b→∞) (−1/b+1) = 1. This converges. ∫₁∞ 1/x dx = lim(b→∞) [ln(x)]₁ᵇ = lim(b→∞) ln(b) = ∞. This diverges. The p-integral ∫₁∞ 1/xᵖ dx converges if p>1, diverges if p≤1. Improper integrals require limits and may converge or diverge — always evaluate the limit explicitly.

Numerical integration approximates definite integrals when antiderivatives are hard or impossible to find. Riemann sum: divide [a,b] into n equal subintervals, approximate each strip as a rectangle. Trapezoid rule: use trapezoids instead of rectangles — error O(h²) where h=(b−a)/n. Simpson's rule: uses parabolic arcs — error O(h⁴), much more accurate. Example: ∫₀¹ e^(−x²) dx ≈ 0.7468 (no elementary antiderivative exists). Gaussian quadrature can achieve very high accuracy with few evaluation points by choosing optimal sampling locations.