Q. 44 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The value is $\mathrm{\pi }$

Step 1. Given Information:

Given double integrals : 

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

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

Step 2. Solution:

$$\begin{gathered} U\sin g Fubini\text{'}s Theorem \\ \int {\int }_R{x}^2\cos \left(xy\right) dA={\int }_0^{\mathrm{\pi }}{\int }_0^1{x}^2\cos \left(xy\right) dy dx \\ Evaluation procedure for the iterated integral we get \\ ={\int }_0^{\mathrm{\pi }}\left({\int }_0^1{x}^2\cos \left(xy\right) dy\right) dx \\ ={\int }_0^{\mathrm{\pi }}\left(x^2{\int }_0^1\cos \left(xy\right) dy\right) dx \end{gathered}$$

$$\begin{gathered} First we solve \\ {\int }_0^1\cos \left(xy\right) dy \\ Put xy=t so we have dy=\frac{dt}{x} \\ When y=1 then t=x \\ When y=0 then t=0 \\ So integral becomes: \\ ={\int }_0^x\frac{\cos t}{x} dt \\ By Fundamental Theorem of Calculus we have \\ =\frac{1}{x}{\left[\sin t\right]}_0^x \\ Evaluation of the inner antiderivative we get \\ =\frac{\sin x}{x} \\ So we have \\ ={\int }_0^{\mathrm{\pi }}{x}^2\frac{\sin x}{x}dx \\ ={\int }_0^{\mathrm{\pi }}x \sin xdx \end{gathered}$$

$$\begin{gathered} By Fundamental Theorem of Calculus we have \\ ={\left[x (-\cos x)\right]}_0^{\mathrm{\pi }}-{\int }_0^{\mathrm{\pi }}-\cos x dx \\ ={\left[x (-\cos x)\right]}_0^{\mathrm{\pi }}+{\left[\sin x\right]}_0^{\mathrm{\pi }} \\ =\left(-\mathrm{\pi } \mathrm{cos\pi }-0\right)+(\mathrm{sin\pi }-0) \\ =\mathrm{\pi } \end{gathered}$$