有限上下限 · 符号或数值验证

带详细步骤的定积分计算器

输入被积函数和有限上下限。工具会使用支持的原函数与微积分基本定理，或给出明确标注的数值计算。

Bounds preconfiguredSubstitution steps shownDiscontinuities checked

专属计算工作台

积分求解器

计算引擎

使用 ^ 表示乘方，函数请使用括号，例如 sin(x)、log(x) 或 exp(2x)。

试试这些例题

From accumulation to a number

定积分是怎样计算的

A definite integral combines an integrand, a variable and two bounds. It returns signed accumulation over an interval and does not include an arbitrary integration constant.

Treat the bounds as part of the problem

The notation ∫ₐᵇf(x)dx accumulates from a to b. Reversing the bounds reverses the sign, and identical bounds give zero.

If the interval crosses a point where the integrand is undefined, ordinary substitution is unsafe. The solver checks for domain failures and reports them.

Apply the Fundamental Theorem

When F′=f on the interval, ∫ₐᵇf(x)dx=F(b)−F(a). For ∫₀²3x²dx, F=x³ and the value is 8.

The integration constant cancels at the two bounds. Keeping the substitution line visible prevents reversing the subtraction or using the original integrand.

Understand signed area

Regions above the axis contribute positively and regions below contribute negatively. A definite integral is therefore not always total geometric area.

If a problem asks for total area or distance, split the interval at sign changes and integrate the required absolute contributions.

Label numerical integration honestly

When no symbolic rule matches, a finite continuous interval can use deterministic Simpson integration and identify the output as an approximation.

Infinite bounds, hidden singularities and improper intervals need convergence analysis, so the current page does not silently approximate them.

A careful definite-integral workflow

A definite integral includes three pieces that must stay together: the integrand, the variable of integration, and the lower and upper bounds. Read ∫ₐᵇf(x)dx as accumulated signed change from a to b. Unlike an indefinite integral, its final output is a value rather than a family of antiderivatives, so the answer does not contain +C. When a symbolic antiderivative is available, use the Fundamental Theorem of Calculus. Find F with F′=f, write the evaluation bar from a to b, and compute F(b)−F(a). For ∫₀²3x²dx, an antiderivative is x³. Substitution gives 2³−0³=8. Keeping the upper-minus-lower line visible is worthwhile because reversing this subtraction is a common error. Any integration constant would occur in both values and cancel. Bounds also determine orientation. Reversing them changes the sign: ∫ᵦᵃf(x)dx=−∫ₐᵇf(x)dx. Equal bounds produce zero because no interval is traversed. These facts provide quick consistency checks before detailed calculation. If a positive function is integrated from a smaller bound to a larger one, a negative result deserves another look at the antiderivative or subtraction. Signed accumulation is not always geometric area. Contributions above the x-axis are positive and contributions below it are negative. If a curve crosses the axis, cancellation can make the integral smaller than the total visible region. A question asking for total area requires locating sign changes, splitting the interval, and adding absolute contributions. A velocity integral gives displacement; total distance similarly requires handling changes of direction. Inspect the interval before using the theorem. A denominator may vanish, a logarithm may receive a nonpositive argument, or a root may leave its real domain. An interval crossing such a point is improper and needs limits and a convergence test. Directly inserting bounds into one antiderivative can produce a plausible but invalid number. CalculusMate checks supported expressions for detected discontinuities and reports the interval instead of silently treating it as proper. When no symbolic integration rule matches, numerical integration can still estimate a finite, continuous interval. The calculator uses a deterministic Simpson procedure and labels the result as approximate. Numerical output is useful for checking scale or solving supported applied problems, but it is different from an exact symbolic form. More subdivisions can reduce ordinary approximation error; they cannot repair a discontinuity or prove convergence at an infinite bound. Use symmetry when appropriate. An odd function integrated over [−a,a] gives zero, while an even function gives twice its integral over [0,a], provided the integral is proper. Units also reveal meaning: integrating a rate measured in units per second over seconds returns units, and integrating height over horizontal distance returns square units. A strong written solution states the antiderivative rule, shows both substitutions, performs upper minus lower, and interprets the sign or units when the context asks for it. The animated steps on this page preserve that sequence. Infinite bounds, Cauchy principal values, and detected singular intervals remain outside the current proper-integral tool because each requires explicit convergence reasoning.

Estimate first, then calculate

Before finding an antiderivative, sketch the sign and rough size of the integrand on the interval. A positive function bounded between two simple heights should produce a value within the corresponding rectangle estimates. This prediction catches many sign, bound, and arithmetic errors. After evaluating, differentiate the proposed antiderivative to confirm it returns the integrand, then check the substitution line separately. For numerical results, compare with a rough geometric estimate and keep the approximation label. Practice reversing bounds, using equal bounds, and splitting an interval at a zero so these structural rules become automatic. When the integrand has a denominator, logarithm, root, or tangent, inspect its domain before calculation. If the interval reaches a singularity or infinity, switch to an improper-integral method with explicit limits rather than treating the endpoint like an ordinary number.

Bounded integral examples

跟着上下限完成代入和相减

∫₀² 3x² dx

Use F=x³ and compute F(2)−F(0)=8.

∫₀^π sin(x) dx

Use F=−cos(x). The upper-minus-lower evaluation produces 2.

∫₀¹ 1/(x²+1) dx

When no symbolic rule matches, the finite continuous interval can be evaluated numerically and labeled approximate.

Proper finite integrals only

Infinite bounds, principal values and intervals crossing a detected singularity are reported as unsupported because they require separate convergence analysis.

FAQ
What is a definite integral?+

It is signed accumulation over an interval, often evaluated as F(b)−F(a).

Why is there no +C?+

The same constant appears at both bounds and cancels in the subtraction.

Does it always equal area?+

It gives signed area. Total geometric area may require splitting the interval and using absolute values.

Can it evaluate improper integrals?+

Not yet. Infinite bounds and detected singularities are reported as unsupported.