Calculators / Calculus 3 / Double Integral Calculator (General Region)

Double Integral Calculator (General Region)

Published on: August 16, 2026

This Double Integral Calculator (General Region) helps you evaluate double integrals over general regions in the plane. The calculator identifies whether the region is Type I or Type II, sets up the correct order of integration, and then works through the integral step by step. This makes it easier to understand the region, the bounds, and the integration process. It is a simple way to check answers, understand the method clearly, and practise multivariable calculus step by step.

It shows the integral setup, identifies the inner and outer variables, classifies the region, and works through the integration step by step.

Enter your problem like f(x,y)=x+y; y:[x^2,x+1]; x:[0,1] or f(x,y)=x*y; x:[y,2]; y:[0,1], or use the ∬ keyboard button.

Step-by-step method

  1. Read the integrand, the inner bounds, and the outer bounds.
  2. Decide whether the setup is Type I or Type II.
  3. Write the iterated integral in the order given by the bounds.
  4. Break the inner integration into smaller algebra steps whenever possible.
  5. Substitute the upper and lower inner bounds separately, then subtract and simplify.
  6. Break the outer integration into smaller algebra steps whenever possible.
  7. Substitute the outer upper and lower bounds separately, then subtract and simplify.
  8. Write the final exact answer, and a decimal approximation when needed.

Formulas:

Type I region setup

∫ba∫g₂( x )g₁( x )f( x, y ) dy dx

Type II region setup

∫dc∫h₂( y )h₁( y )f( x, y ) dx dy

Example 1: f(x,y)=x+y; y:[x^2,x+1]; x:[0,1]

Step 1A - Identify the integrand.

In this problem: Read the function being integrated.

f( x, y ) = x + y

Step 1B - Identify the inner bounds.

In this problem: Inner variable: y. Inner bounds: x^2 ≤ y ≤ x + 1.

x2 ≤ y ≤ 1 + x

Step 1C - Identify the outer bounds.

In this problem: Outer variable: x. Outer bounds: 0 ≤ x ≤ 1.

0 ≤ x ≤ 1

Step 1D - Classify the region.

In this problem: Type I. R = { (x, y) : 0 ≤ x ≤ 1, x^2 ≤ y ≤ x + 1 }. This is Type I because y is between lower and upper curves written as functions of x, while x has constant limits.

I = ∫10∫1 + xx2x + y dy dx

Step 2A - Set up the inner integral.

In this problem: Integrate first with respect to y.

I₁ = ∫1 + xx2x + y dy

Step 2B - Split the inner integral into simpler pieces.

In this problem: Use linearity of integration to integrate one term at a time with respect to y.

∫ x + y dy = ∫ x dy + ∫ y dy

Step 2C.1 - Integrate term 1.

In this problem: Work only on the term x.

∫ x dy = x·y

Step 2C.2 - Integrate term 2.

In this problem: Work only on the term y.

∫ y dy =
y2
2

Step 2D - Combine the inner antiderivative pieces.

In this problem: Add the antiderivatives of all terms.

F = xy +
y2
2

Step 2E - Substitute the upper inner bound.

In this problem: Use the upper inner bound y = x + 1.

F = xy +
1
2
y2
=
1
2
(1 + x)2 + x(1 + x)
=
1
2
+
4x
2
+
3x2
2
=
1 + 3x2 + 4x
2

Step 2F - Substitute the lower inner bound.

In this problem: Use the lower inner bound y = x^2.

F = xy +
1
2
y2
= x3 +
1
2
x4
=
2x3
2
+
x4
2
=
x4 + 2x3
2

Step 2G - Subtract upper minus lower.

In this problem: Now subtract the lower result from the upper result.

I₁ =
1
2
+ 2x +
3
2
x2 − x3 +
1
2
x4
=
1
2
−
2x3
2
+
4x
2
−
x4
2
+
3x2
2
=
1 − x4 − 2x3 + 3x2 + 4x
2

Step 2H - Simplify the inner result.

In this problem: This leaves an expression in the outer variable x.

I₁ =
1 − x4 − 2x3 + 3x2 + 4x
2

Step 3A - Set up the outer integral.

In this problem: Now integrate the inner result with respect to x.

I = ∫10
1 − x4 − 2x3 + 3x2 + 4x
2
 dx

Step 3B - Split the outer integral into simpler pieces.

In this problem: Use linearity of integration to integrate one term at a time with respect to x.

∫
1 − x4 − 2x3 + 3x2 + 4x
2
dx = ∫
1
2
dx + ∫ −x3 dx + ∫ 2x dx + ∫
−x4
2
dx + ∫
3x2
2
dx

Step 3C.1 - Integrate term 1.

In this problem: Work only on the term 1/2.

∫
1
2
dx =
1
2
·x

Step 3C.2 - Integrate term 2.

In this problem: Work only on the term -x^3.

∫ −x3 dx = -1·∫ x3 dx
= -1·
x4
4
=
−x4
4

Step 3C.3 - Integrate term 3.

In this problem: Work only on the term 2*x.

∫ 2x dx = 2·∫ x dx
= 2·
x2
2
= x2

Step 3C.4 - Integrate term 4.

In this problem: Work only on the term -x^4/2.

∫
−x4
2
dx =
-1
2
·∫ x4 dx
=
-1
2
·
x5
5
=
−x5
10

Step 3C.5 - Integrate term 5.

In this problem: Work only on the term 3*x^2/2.

∫
3x2
2
dx =
3
2
·∫ x2 dx
=
3
2
·
x3
3
=
x3
2

Step 3D - Combine the outer antiderivative pieces.

In this problem: Add the antiderivatives of all terms.

G =
x
2
−
x4
4
+ x2 −
x5
10
+
x3
2

Step 3E - Substitute the upper outer bound.

In this problem: Use the upper outer bound x = 1.

G =
1
2
x −
1
4
x4 + x2 −
1
10
x5 +
1
2
x3
=
33
20

Step 3F - Substitute the lower outer bound.

In this problem: Use the lower outer bound x = 0.

G =
1
2
x −
1
4
x4 + x2 −
1
10
x5 +
1
2
x3
= 0

Step 3G - Subtract upper minus lower.

In this problem: Now subtract the lower result from the upper result.

I =
33
20
− 0

Step 3H - Write the final answer.

In this problem: This is the value of the double integral over the given region.

I =
33
20

Final answer: 33/20

Example 2: f(x,y)=x*y; x:[y,2]; y:[0,1]

Step 1A - Identify the integrand.

In this problem: Read the function being integrated.

f( x, y ) = xy

Step 1B - Identify the inner bounds.

In this problem: Inner variable: x. Inner bounds: y ≤ x ≤ 2.

y ≤ x ≤ 2

Step 1C - Identify the outer bounds.

In this problem: Outer variable: y. Outer bounds: 0 ≤ y ≤ 1.

0 ≤ y ≤ 1

Step 1D - Classify the region.

In this problem: Type II. R = { (x, y) : 0 ≤ y ≤ 1, y ≤ x ≤ 2 }. This is Type II because x is between left and right curves written as functions of y, while y has constant limits.

I = ∫10∫2yxy dx dy

Step 2A - Set up the inner integral.

In this problem: Integrate first with respect to x.

I₁ = ∫2yxy dx

Step 2B - Find the inner antiderivative.

In this problem: Integrate with respect to x.

∫ xy dx = y·∫ x dx
= y·
x2
2
=
yx2
2
F =
yx2
2

Step 2E - Substitute the upper inner bound.

In this problem: Use the upper inner bound x = 2.

F =
yx2
2
= 2y

Step 2F - Substitute the lower inner bound.

In this problem: Use the lower inner bound x = y.

F =
yx2
2
=
y3
2

Step 2G - Subtract upper minus lower.

In this problem: Now subtract the lower result from the upper result.

I₁ = 2y −
y3
2
=
4y
2
−
y3
2
=
−y3 + 4y
2

Step 2H - Simplify the inner result.

In this problem: This leaves an expression in the outer variable y.

I₁ =
y(4 − y2)
2

Step 3A - Set up the outer integral.

In this problem: Now integrate the inner result with respect to y.

I = ∫10
y(4 − y2)
2
 dy

Step 3B - Split the outer integral into simpler pieces.

In this problem: Use linearity of integration to integrate one term at a time with respect to y.

∫
y(4 − y2)
2
dy = ∫ 2y dy + ∫
−y3
2
dy

Step 3C.1 - Integrate term 1.

In this problem: Work only on the term 2*y.

∫ 2y dy = 2·∫ y dy
= 2·
y2
2
= y2

Step 3C.2 - Integrate term 2.

In this problem: Work only on the term -y^3/2.

∫
−y3
2
dy =
-1
2
·∫ y3 dy
=
-1
2
·
y4
4
=
−y4
8

Step 3D - Combine the outer antiderivative pieces.

In this problem: Add the antiderivatives of all terms.

G = y2 −
y4
8

Step 3E - Substitute the upper outer bound.

In this problem: Use the upper outer bound y = 1.

G = y2 −
1
8
y4
=
7
8

Step 3F - Substitute the lower outer bound.

In this problem: Use the lower outer bound y = 0.

G = y2 −
1
8
y4
= 0

Step 3G - Subtract upper minus lower.

In this problem: Now subtract the lower result from the upper result.

I =
7
8
− 0

Step 3H - Write the final answer.

In this problem: This is the value of the double integral over the given region.

I =
7
8

Final answer: 7/8