Double Integral Calculator (Rectangular Region)

Published on: July 21, 2026
Final Answer: Free • Full Steps: Pro

This Double Integral Calculator evaluates \(\iint_R f(x,y)\,dA\) over a rectangle. It integrates with respect to \(x\) first (holding \(y\) constant), then with respect to \(y\), showing each integration and evaluation step.

Step-by-step method

  1. Set up the integrand and the rectangular limits.
  2. Integrate with respect to x first (inner integral), treating y as constant.
  3. Evaluate the inner integral at the x-limits.
  4. Integrate the result with respect to y (outer integral).
  5. Evaluate at the y-limits to get the final value.

Formula:

\(\iint_R f\,dA = \int_{c}^{d}\!\int_{a}^{b} f(x,y)\,dx\,dy\)

Example 1:

\(\int_{0}^{2}\!\int_{0}^{1} x y\,dx\,dy\)

Step 1 - Set up the integrand and the rectangular limits.

In this problem: We integrate \(f = x y\) over \(x \in [0, 1]\), \(y \in [0, 2]\).

\(\int_{0}^{2}\!\int_{0}^{1} x y\,dx\,dy\)

Step 2 - Integrate with respect to x first (inner integral), treating y as constant.

In this problem: Treating \(y\) as constant, an antiderivative in \(x\) is \(\frac{x^{2} y}{2}\).

\(\int x y\,dx = \frac{x^{2} y}{2} + C\)

Step 3 - Evaluate the inner integral at the x-limits.

In this problem: Evaluating from \(x = 0\) to \(x = 1\) gives \(\frac{y}{2}\).

\(\int_{0}^{1} x y\,dx = \frac{y}{2}\)

Step 4 - Integrate the result with respect to y (outer integral).

In this problem: Now integrate \(\frac{y}{2}\) with respect to \(y\): antiderivative \(\frac{y^{2}}{4}\).

\(\int \frac{y}{2}\,dy = \frac{y^{2}}{4} + C\)

Step 5 - Evaluate at the y-limits to get the final value.

In this problem: Evaluating from \(y = 0\) to \(y = 2\) gives \(1\).

\(\iint_R f\,dA = 1\)

Final answer:

\(\iint_R f\,dA = 1\)

Example 2:

\(\int_{1}^{2}\!\int_{0}^{1} x^{2} + y\,dx\,dy\)

Step 1 - Set up the integrand and the rectangular limits.

In this problem: We integrate \(f = x^{2} + y\) over \(x \in [0, 1]\), \(y \in [1, 2]\).

\(\int_{1}^{2}\!\int_{0}^{1} x^{2} + y\,dx\,dy\)

Step 2 - Integrate with respect to x first (inner integral), treating y as constant.

In this problem: Treating \(y\) as constant, an antiderivative in \(x\) is \(\frac{x^{3}}{3} + x y\).

\(\int x^{2} + y\,dx = \frac{x^{3}}{3} + x y + C\)

Step 3 - Evaluate the inner integral at the x-limits.

In this problem: Evaluating from \(x = 0\) to \(x = 1\) gives \(y + \frac{1}{3}\).

\(\int_{0}^{1} x^{2} + y\,dx = y + \frac{1}{3}\)

Step 4 - Integrate the result with respect to y (outer integral).

In this problem: Now integrate \(y + \frac{1}{3}\) with respect to \(y\): antiderivative \(\frac{y^{2}}{2} + \frac{y}{3}\).

\(\int y + \frac{1}{3}\,dy = \frac{y^{2}}{2} + \frac{y}{3} + C\)

Step 5 - Evaluate at the y-limits to get the final value.

In this problem: Evaluating from \(y = 1\) to \(y = 2\) gives \(\frac{11}{6} \approx 1.83\).

\(\iint_R f\,dA = \frac{11}{6} \approx 1.83\)

Final answer:

\(\iint_R f\,dA = \frac{11}{6} \approx 1.83\)