Double Integral Calculator (General 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 region where \(y\) runs between two curves \(g_1(x)\) and \(g_2(x)\). It integrates in \(y\) first, then in \(x\), showing each step.

Step-by-step method

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

Formula:

\(\iint_R f\,dA = \int_{a}^{b}\!\int_{g_{1}(x)}^{g_{2}(x)} f(x,y)\,dy\,dx\)

Example 1:

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

Step 1 - Set up the integrand and the region (y between two curves).

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

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

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

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

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

Step 3 - Evaluate the inner integral at the y-curves.

In this problem: Evaluating from \(y = 0\) to \(y = x\) gives \(\frac{x^{3}}{2}\).

\(\int_{0}^{x} x y\,dy = \frac{x^{3}}{2}\)

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

In this problem: Now integrate \(\frac{x^{3}}{2}\) with respect to \(x\).

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

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

In this problem: Evaluating from \(x = 0\) to \(x = 1\) gives \(\frac{1}{8} \approx 0.12\).

\(\iint_R f\,dA = \frac{1}{8} \approx 0.12\)

Final answer:

\(\iint_R f\,dA = \frac{1}{8} \approx 0.12\)

Example 2:

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

Step 1 - Set up the integrand and the region (y between two curves).

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

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

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

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

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

Step 3 - Evaluate the inner integral at the y-curves.

In this problem: Evaluating from \(y = x^{2}\) to \(y = x\) gives \(\frac{x^{2} \left(- x^{2} - 2 x + 3\right)}{2}\).

\(\int_{x^{2}}^{x} x + y\,dy = \frac{x^{2} \left(- x^{2} - 2 x + 3\right)}{2}\)

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

In this problem: Now integrate \(\frac{x^{2} \left(- x^{2} - 2 x + 3\right)}{2}\) with respect to \(x\).

\(\int_{0}^{1} \frac{x^{2} \left(- x^{2} - 2 x + 3\right)}{2}\,dx\)

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

In this problem: Evaluating from \(x = 0\) to \(x = 1\) gives \(\frac{3}{20}\).

\(\iint_R f\,dA = \frac{3}{20}\)

Final answer:

\(\iint_R f\,dA = \frac{3}{20}\)