Iterated Integrals — Question 5

Question 5 – Solution

Strategy. Integrate the positive $y$-factor first, then split the remaining $x$-integral where $x-1$ changes sign.

Step 1: Inner integral Using $dy$ first, $$\begin{align*} \int_{-1}^{3}\int_0^2|x-1|(1+y)\,dy\,dx &=\int_{-1}^{3}|x-1|\left[y+\frac{y^2}{2}\right]_0^2dx\\ &=4\int_{-1}^{3}|x-1|\,dx. \end{align*}$$

Step 2: Absolute-value split $$\begin{align*} \int_{-1}^{3}|x-1|\,dx &=\int_{-1}^{1}(1-x)\,dx+\int_1^3(x-1)\,dx\\ &=2+2=4. \end{align*}$$ Therefore $$\boxed{\iint_R|x-1|(1+y)\,dA=16}.$$

Step 3: Average and check The rectangle has area $4\cdot 2=8$, so $$\boxed{f_{\mathrm{avg}}=\frac{16}{8}=2}.$$ The two $x$-pieces are mirror images about $x=1$, and the nonnegative integrand confirms the positive sign.