Double integrals: choose the region and integrate in order
A double integral adds contributions over an area. In an iterated integral, the differential nearest the integrand tells you which variable to integrate first. While integrating with respect to y, treat x as a constant. For a rectangular region the limits can be constant; for a triangle one boundary usually depends on the other variable. Sketching or describing the boundaries first prevents a correct antiderivative being evaluated over the wrong region.
Worked practice example
Evaluate ∫ from x = 0 to 2 ∫ from y = 0 to 1 of (x + 2y) dy dx. The region is the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
| Step | Working |
|---|---|
| Inner integral | ∫(x + 2y) dy = xy + y². |
| Apply y limits | [xy + y²] from 0 to 1 = x + 1. |
| Outer integral | ∫ from 0 to 2 (x + 1) dx = [x²/2 + x] from 0 to 2. |
| Result | 2 + 2 = 4. |
Check your method
The integrand is nonnegative on the whole rectangle, so a negative answer would signal an error. Reversing the integration order over this same rectangle gives the same result. Do not add a new arbitrary constant after each definite integration.
Try it yourself
Evaluate ∫ from 0 to 1 ∫ from 0 to 2 of 3xy dy dx.
Show the answer and method
First integrate in y: (3x/2)y² gives 6x after applying 0 and 2. Then ∫ from 0 to 1 6x dx = 3.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.