Iterated Integrals — Question 8

Question 8 – Solution

Strategy. In the inner integral, the factor depending only on the outer variable is constant and can be pulled outside.

Step 1: Factorization proof Integrating in $y$ first gives $$\begin{align*} \int_a^b\int_c^d p(x)q(y)\,dy\,dx &=\int_a^b p(x)\left(\int_c^d q(y)\,dy\right)dx\\ &=\left(\int_a^b p(x)\,dx\right) \left(\int_c^d q(y)\,dy\right). \end{align*}$$ The bracketed $q$-integral is a number, so it is constant in the outer $x$-integration.

Step 2: Other order Reversing the order first produces the constant $\int_a^b p(x)\,dx$ and then the $q$-integral. Continuity on the closed rectangle guarantees Fubini’s theorem applies, so both forms represent the same double integral.

Step 3: Application Here $$\int_{-1}^{1}(1+x^2)\,dx =\left[x+\frac{x^3}{3}\right]_{-1}^{1}=\frac 83$$ and $$\int_{-\pi/2}^{\pi/2}\cos y\,dy=[\sin y]_{-\pi/2}^{\pi/2}=2.$$ Therefore $$\boxed{\iint_R(1+x^2)\cos y\,dA=\frac{16}{3}}.$$ Both factors are nonnegative on their intervals, verifying the sign.