Finding the Value of a Double Integral Using Fubini’s Theorem

Find the value of \(\displaystyle\iint\limits_{\kern-3ptR}(2x+y)\,{\it dA}\) if \(R\) is the closed rectangular region defined by \(1\leq x\leq 5\) and \(0\leq y\leq 4\).

Solution Using Fubini’s Theorem, the double integral \(\iint\limits_{\kern-3ptR}(2x+y)\,{\it dA}\) equals either of the iterated integrals \[ \begin{equation} \displaystyle\int_{1}^{5}\left[ \int_{0}^{4}(2x+y)\,{dy}\right]\! {dx}\qquad \hbox{or} \qquad \displaystyle\int_{0}^{4} \left[ \int_{1}^{5}(2x+y)\,{dx}\right]\! {dy} \tag{1} \end{equation} \]

908

Below we find the integral on the left. \[ \begin{eqnarray*} \displaystyle\iint\limits_{\kern-3ptR}(2x+y)\,{\it dA}&=&\int_{1}^{5}\left[ \int_{0}^{4}(2x+y)\,{\it dy}\right]\! {\it dx} \underset{\underset{\underset{{\color{#0066A7}{\hbox{with respect to $y$}}}}{\color{#0066A7}{\hbox{Integrate partially }}}}{\color{#0066A7}{\uparrow }}}{=}\int_{1}^{5} \left[2xy + \dfrac{y^{2}}{2}\right] _{0}^{4}\,{\it dx} \\ &=& \int^5_1 (8x+8)\,{\it dx} = \big[4x^2+8x\big]^5_1 = 140-12=128 \end{eqnarray*} \]