Iterated Integrals — Question 4

Question 4 – Solution

Strategy. Read an iterated integral from the inside out: the inner differential determines both the variable of integration and which pair of bounds belongs to it.

Step 1: Diagnose The inner differential is $dx$, so $x$ runs from $0$ to $2$ while $y$ is held constant. The student incorrectly used the outer $y$-bounds, $-1$ and $1$, as $x$-bounds.

Step 2: Evaluate as written $$\begin{align*} I&=\int_{-1}^{1}\left[\frac{x^2}{2}-2yx\right]_0^2dy =\int_{-1}^{1}(2-4y)\,dy\\ &=[2y-2y^2]_{-1}^{1}=\boxed{4}. \end{align*}$$ The odd contribution $-4y$ cancels over $[-1,1]$.

Step 3: Reverse correctly Constant rectangular bounds reverse to $$I=\int_0^2\int_{-1}^{1}(x-2y)\,dy\,dx.$$ Then $$\begin{align*} I&=\int_0^2[xy-y^2]_{-1}^{1}\,dx =\int_0^2 2x\,dx=[x^2]_0^2=4. \end{align*}$$ The agreement verifies both the order reversal and the association of each bound pair with its variable.