Iterated Integrals — Question 1

Question 1 – Solution

Strategy. Write both constant-bound orders explicitly and simplify the inner integral before performing the outer one.

Step 1: Integrate in $y$ first $$\begin{align*} \int_0^2\int_{-1}^{1}(3x^2y-2x+4)\,dy\,dx &=\int_0^2\left[\frac 32x^2y^2-2xy+4y\right]_{-1}^{1}dx\\ &=\int_0^2(-4x+8)\,dx\\ &=\left[-2x^2+8x\right]_0^2=8. \end{align*}$$ The $3x^2y$ term vanishes because it is odd in $y$ over a symmetric interval.

Step 2: Integrate in $x$ first $$\begin{align*} \int_{-1}^{1}\int_0^2(3x^2y-2x+4)\,dx\,dy &=\int_{-1}^{1}\left[x^3y-x^2+4x\right]_0^2dy\\ &=\int_{-1}^{1}(8y+4)\,dy=8. \end{align*}$$ Thus $$\boxed{\iint_R f\,dA=8}.$$

Verification The polynomial is continuous on the closed rectangle, so Fubini’s theorem permits either order. In the second order the odd term $8y$ cancels, leaving constant height $4$ across a $y$-interval of length $2$, again giving $8$.