3. change the order of integration so that you can evaluate the integrals (and then evaluate…

3. change the order of integration so that you can evaluate the integrals (and then evaluate them!)\n(a)\n$\\int_{0}^{1}\\int_{0}^{2}xe^{3xy}dxdy$\n(b)\n$\\int_{0}^{1}\\int_{0}^{y^{2}}\\cos(\\pi xy)dydx$\n

3. change the order of integration so that you can evaluate the integrals (and then evaluate them!)\n(a)\n$\\int_{0}^{1}\\int_{0}^{2}xe^{3xy}dxdy$\n(b)\n$\\int_{0}^{1}\\int_{0}^{y^{2}}\\cos(\\pi xy)dydx$\n

Answer

(a)

Explanation:

Step1: Determine the region of integration

The original limits are (0\leq x\leq1) and (1\leq y\leq2). The region (R) is a rectangle. When changing the order of integration (from (dxdy) to (dydx)), the limits for (y) remain (1\leq y\leq2) and for (x) remain (0\leq x\leq1) since it's a rectangle. The integral (\int_{0}^{1}\int_{1}^{2}xe^{3xy}dxdy) after changing the order becomes (\int_{1}^{2}\int_{0}^{1}xe^{3xy}dydx)

Step2: Integrate with respect to (y) first

(\int_{0}^{1}xe^{3xy}dy). Let (u = 3xy), then (du=3xdy). When (y = 0), (u = 0); when (y = 1), (u=3x). (\int_{0}^{1}xe^{3xy}dy=\frac{1}{3}\int_{0}^{3x}e^{u}du=\frac{1}{3}(e^{3x}-1))

Step3: Integrate the result with respect to (x)

(\int_{1}^{2}\frac{1}{3}(e^{3x}-1)dx=\frac{1}{9}e^{3x}-\frac{1}{3}x\big|_{1}^{2}) [ \begin{align*} &=\left(\frac{1}{9}e^{6}-\frac{2}{3}\right)-\left(\frac{1}{9}e^{3}-\frac{1}{3}\right)\ &=\frac{1}{9}(e^{6}-e^{3})-\frac{1}{3} \end{align*} ]

(b)

Explanation:

Step1: Determine the region of integration

The original limits are (0\leq x\leq1) and (x^{2}\leq y\leq1). The curves (y = x^{2}) ((x=\sqrt{y})) and (x = 0,y = 0,y = 1) bound the region. When changing the order of integration (from (dydx) to (dxdy)), the limits for (y) are (0\leq y\leq1) and for (x) are (0\leq x\leq\sqrt{y}) The integral (\int_{0}^{1}\int_{x^{2}}^{1}y^{2}\cos(\pi xy)dydx) becomes (\int_{0}^{1}\int_{0}^{\sqrt{y}}y^{2}\cos(\pi xy)dxdy)

Step2: Integrate with respect to (x) first

(\int_{0}^{\sqrt{y}}y^{2}\cos(\pi xy)dx). Let (u=\pi xy), then (du=\pi ydx). When (x = 0), (u = 0); when (x=\sqrt{y}), (u=\pi y^{3/2}) (\int_{0}^{\sqrt{y}}y^{2}\cos(\pi xy)dx=\frac{y}{\pi}\int_{0}^{\pi y^{3/2}}\cos(u)du=\frac{y}{\pi}\sin(\pi y^{3/2}))

Step3: Integrate the result with respect to (y)

Let (t=\pi y^{3/2}), then (dt=\frac{3\pi}{2}y^{1/2}dy) or (y dy=\frac{2}{3\pi}dt) (\int_{0}^{1}\frac{y}{\pi}\sin(\pi y^{3/2})dy=\frac{2}{3\pi^{2}}\int_{0}^{\pi}\sin(t)dt) [ \begin{align*} &=\frac{2}{3\pi^{2}}(-\cos t)\big|_{0}^{\pi}\ &=\frac{4}{3\pi^{2}} \end{align*} ]

Answer:

(a) (\frac{1}{9}(e^{6}-e^{3})-\frac{1}{3})

(b) (\frac{4}{3\pi^{2}})