Evaluate the double integral of (x/8)(x² + xy) over the region 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

Step-by-step solution

1. 1

   Expand the integrand before doing anything else. A product like $\frac{x}{8}(x^2 + xy)$ is easier to integrate once it is a sum of monomials:

   $\frac{x}{8}(x^2 + xy) = \frac{x^3}{8} + \frac{x^2 y}{8}$

   Now each term is a simple power of $x$ times a simple power of $y$, so both integrations are elementary.

2. 2

   Integrate in y with x held constant. The factor $x^3/8$ is constant in $y$, while $x^2y/8$ needs the $y$-power rule:

   $\int_0^2 \left(\frac{x^3}{8} + \frac{x^2 y}{8}\right) dy = \left[\frac{x^3 y}{8} + \frac{x^2 y^2}{16}\right]_0^2$

3. 3

   Substitute the y-limits. At $y = 2$ (the $y = 0$ end contributes nothing):

   $\frac{2x^3}{8} + \frac{4x^2}{16} = \frac{x^3}{4} + \frac{x^2}{4}$

   Notice the $y^2$ term produces $4$, not $2$ — squaring the upper limit is where an arithmetic error usually enters.

4. 4

   Integrate the result in x. Pull out the common $\tfrac14$:

   $\frac{1}{4}\int_0^2 (x^3 + x^2)\,dx = \frac{1}{4}\left[\frac{x^4}{4} + \frac{x^3}{3}\right]_0^2 = \frac{1}{4}\left(4 + \frac{8}{3}\right)$

5. 5

   Finish the arithmetic and sanity-check the size.

   $\frac{1}{4} \cdot \frac{20}{3} = \frac{5}{3} \approx 1.6667$

   Quadrature over the square agrees to eight digits. As a plausibility check, the integrand ranges from $0$ to $\frac{2}{8}(4 + 4) = 2$ on a region of area $4$, so a value of $\tfrac53$ is comfortably inside the possible range. (The $1/8$ is exactly the constant that would make $\frac{x}{8}(x+y)$ a probability density on this square — this integrand is that density times $x$.)