Integration is the mathematics of accumulation: adding up countless tiny pieces to find a total. Area under a curve, distance travelled from a speed record, volume built from slices, charge accumulated from a current — all of these are integrals. The integral calculator evaluates integrals of a function you supply, both indefinite integrals (the general antiderivative, the reverse of differentiation) and definite integrals between two bounds (a single number, the net accumulation). It applies the standard integration rules accurately, sparing you the algebraic grind and the slips that come with it.
Quick Answer
The workhorse rule mirrors the power rule for derivatives: ∫ x^n dx = x^(n+1) / (n + 1) + C, valid for n ≠ −1. For a definite integral, evaluate the antiderivative at the top bound and subtract its value at the bottom bound: ∫ from a to b = F(b) − F(a).
Integration as Accumulation — and Reverse Differentiation
Picture the area under a curve chopped into thin vertical strips. Each strip is roughly a rectangle; add them all and let the strips shrink toward zero width, and the sum converges to the exact area. That limit of sums is the definite integral, and its symbol ∫ is an elongated S for "sum". The miracle — the Fundamental Theorem of Calculus — is that this geometric problem is solved by an algebraic one: if you can find a function whose derivative is the integrand, subtracting its values at the two bounds gives the area exactly.
That is why indefinite and definite integrals feel like cousins. The indefinite integral ∫ f(x) dx asks "what differentiates to f(x)?" and its answer is a family of functions, differing only by a constant C, because constants vanish on differentiation. The definite integral asks "how much accumulated between a and b?" and returns a number — interpretable as signed area, where area below the axis counts negative. Units follow the product: integrate a rate in litres per second over time in seconds and the answer is in litres.
How to Use the Integral Calculator
- Enter the integrand — the function being integrated — in terms of x, using ^ for powers and standard function names.
- Decide which kind of integral you need: indefinite for the antiderivative formula, definite for a numeric accumulation.
- For a definite integral, enter the lower bound first and the upper bound second; the order matters, because reversing it flips the sign.
- Run the calculation and read the result.
- For an indefinite answer, remember the + C is part of the result even when it is left implicit.
- Verify by differentiating the answer: you should recover your original integrand exactly.
Worked Example: Integrating 3x² + 2x
Integrate term by term. For 3x², raise the power and divide: 3 · x³/3 = x³. For 2x, the same rule gives 2 · x²/2 = x². So ∫ (3x² + 2x) dx = x³ + x² + C. Differentiate to check: d/dx of x³ + x² is 3x² + 2x — the integrand, as required.
Now make it definite. ∫ from 0 to 2 of 3x² dx uses the antiderivative x³: F(2) − F(0) = 8 − 0 = 8, the area under y = 3x² between x = 0 and x = 2. A second one: ∫ from 1 to 3 of x dx = [x²/2] from 1 to 3 = 9/2 − 1/2 = 4, which geometry confirms — the region is a trapezoid that a moment's sketch shows has area 4.
- ∫ 1 dx = x + C
- ∫ (1/x) dx = ln|x| + C — the exception to the power rule
- ∫ e^x dx = e^x + C
- ∫ sin(x) dx = −cos(x) + C; ∫ cos(x) dx = sin(x) + C
- ∫ from 0 to π of sin(x) dx = 2
Displacement, Distance, and Average Value
Two refinements separate routine answers from right ones in applied problems. First, integrating a velocity gives displacement — the net change of position — not the total distance travelled. If the velocity changes sign, the object backtracked, and the parts of the integral below the time axis subtract. Total distance requires integrating the speed, |v(t)|, splitting the integral wherever the velocity crosses zero. Second, a definite integral hides an average: the average value of f between a and b is the integral divided by (b − a). For the earlier example, ∫ from 1 to 3 of x dx = 4, so the average value of x over [1, 3] is 4/2 = 2 — plausibly, the value the function takes at the interval's midpoint, x = 2. Power grids billing on average load, and engineers quoting mean pressure, are using exactly this quotient.
The Constant of Integration and Other Traps
Forgetting + C is the signature indefinite-integral error: two antiderivatives that differ by a constant have the same derivative, so the family is not complete without it, and applied problems pin C down with an initial condition. The power rule's exception catches people too — it divides by n + 1, so at n = −1 it divides by zero; that integral is ln|x| instead. Definite integrals have their own traps: swapping the bounds changes the sign, and an answer can legitimately come out negative when the curve dips below the axis, which is correct as net accumulation but wrong if the question asked for total geometric area — there you integrate the absolute value or split at the crossings. Finally, keep constants honest: a factor of 3 in the integrand stays a factor of 3 outside, and dropping it is the quietest way to lose a third of your answer.
Round Out Your Calculus Toolkit
Every integral you compute can be checked with the Derivative Calculator, since differentiation undoes integration — the fastest verification in all of calculus. When your problem is specifically a bounded area or accumulation, the Definite Integral Calculator focuses on that case alone. Both sit in the math collection alongside this tool.