Question

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.

$$\begin{gathered} \int {\int }_R{x}^2\cos \left(xy\right)dA, \\ whereR=\left\{\right(x,y\left)\right|0\le x\le \pi and0\le y\le 1\} \end{gathered}$$

Short answer

The value is$\mathrm{\pi }$

Step-by-step solution

Step 1. Given Information:

Given double integrals :

$$\begin{gathered} \int {\int }_R{x}^2\cos \left(xy\right)dA, \\ whereR=\left\{\right(x,y\left)\right|0\le x\le \pi and0\le y\le 1\} \end{gathered}$$

We want to evaluate each of the double integrals as iterated integrals.

Step 2. Solution:

$$\begin{gathered} U\sin gFubini\text{'}sTheorem \\ \int {\int }_R{x}^2\cos \left(xy\right)dA={\int }_0^{\mathrm{\pi }}{\int }_0^1{x}^2\cos \left(xy\right)dydx \\ Evaluationprocedurefortheiteratedintegralweget \\ ={\int }_0^{\mathrm{\pi }}\left({\int }_0^1{x}^2\cos \left(xy\right)dy\right)dx \\ ={\int }_0^{\mathrm{\pi }}\left(x^2{\int }_0^1\cos \left(xy\right)dy\right)dx \end{gathered}$$

$$\begin{gathered} Firstwesolve \\ {\int }_0^1\cos \left(xy\right)dy \\ Putxy=tsowehavedy=\frac{dt}{x} \\ Wheny=1thent=x \\ Wheny=0thent=0 \\ Sointegralbecomes: \\ ={\int }_0^x\frac{\cos t}{x}dt \\ ByFundamentalTheoremofCalculuswehave \\ =\frac{1}{x}{\left[\sin t\right]}_0^x \\ Evaluationoftheinnerantiderivativeweget \\ =\frac{\sin x}{x} \\ Sowehave \\ ={\int }_0^{\mathrm{\pi }}{x}^2\frac{\sin x}{x}dx \\ ={\int }_0^{\mathrm{\pi }}x\sin xdx \end{gathered}$$

$$\begin{gathered} ByFundamentalTheoremofCalculuswehave \\ ={\left[x(-\cos x)\right]}_0^{\mathrm{\pi }}-{\int }_0^{\mathrm{\pi }}-\cos xdx \\ ={\left[x(-\cos x)\right]}_0^{\mathrm{\pi }}+{\left[\sin x\right]}_0^{\mathrm{\pi }} \\ =\left(-\mathrm{\pi }\mathrm{cos\pi }-0\right)+(\mathrm{sin\pi }-0) \\ =\mathrm{\pi } \end{gathered}$$