Calcolatrice di integrali doppi

Questa calcolatrice calcola ∬ f(x, y) dA su un rettangolo a ≤ x ≤ b, c ≤ y ≤ d per un polinomio f come x + y o 6x²y. Risolve l'integrale interno e poi quello esterno, dà il risultato esatto e il valore medio di f, e verifica il risultato integrando nell'altro ordine.

Risultato

Inserisci i valori e premi Calcola per vedere il risultato.

Formula

∬Rf(x,y) dA=∫ab(∫cdf(x,y) dy)dx\iint_R f(x, y)\, dA = \int_a^b \left( \int_c^d f(x, y)\, dy \right) dx

An iterated integral: integrate with respect to y holding x fixed, then integrate the result with respect to x.

∫ab∫cdf dy dx=∫cd∫abf dx dy\int_a^b \int_c^d f\, dy\, dx = \int_c^d \int_a^b f\, dx\, dy

Fubini’s theorem: over a rectangle, both orders give the same answer.

Come usare questa calcolatrice

  1. Type f(x, y) as a polynomial in x and y, such as x*y^2 or x^2 + 3y (xy also means x × y).
  2. Enter the limits: x from a to b and y from c to d.
  3. Choose the order of integration and press Calculate.

Esempi svolti

Easy: ∬ (x + y) dA over 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

Inner integral: ∫₀² (x + y) dy = [xy + y²/2] from 0 to 2 = 2x + 2.

Outer integral: ∫₀¹ (2x + 2) dx = [x² + 2x] from 0 to 1 = 3.

The rectangle has area 2, so the average value of f is 3/2.

Medium: ∬ xy² dA, integrating x first

Inner integral: ∫₀² xy² dx = [x²y²/2] from 0 to 2 = 2y².

Outer integral: ∫₁³ 2y² dy = [2y³/3] from 1 to 3 = 18 − 2/3 = 52/3.

Area 4, so the average value is 13/3.

Hard: negative limits

Inner integral: ∫₀² (x² + 3y) dy = [x²y + 3y²/2] from 0 to 2 = 2x² + 6.

Outer integral: ∫₋₁¹ (2x² + 6) dx = [2x³/3 + 6x] from −1 to 1 = 4/3 + 12 = 40/3.

Area 4, so the average value is 10/3.

Errori comuni

  • Treating the outer variable as a variable in the inner integral: hold it constant.
  • Using the wrong limits: in dy dx, the inner limits are the y-limits.
  • Forgetting to subtract the lower limit when it isn’t 0.
  • Mixing up the order: the differential written first (dy in dy dx) is integrated first.

Domande frequenti

How do you evaluate a double integral?

As two single integrals. Integrate with respect to the inner variable, treating the other as a constant, substitute the inner limits, then integrate the result over the outer limits.

Does the order of integration matter?

Not over a rectangle with a polynomial: Fubini’s theorem gives the same answer either way, and the calculator checks that. For non-rectangular regions the limits change when you switch order.

What does a double integral represent?

For f ≥ 0, the volume under the surface z = f(x, y) above the rectangle. Dividing by the rectangle’s area gives the average value of f.

Which functions and regions are supported?

Polynomials in x and y (total degree up to 12) over rectangles a ≤ x ≤ b, c ≤ y ≤ d. Answers are exact fractions.