1. evaluate the following integrals:\n(a) $$ \\int_{0}^{2} \\int_{0}^{3} d x d y $$.\n(b) $$ \\int_{0}^{2}…

1. evaluate the following integrals:\n(a) $$ \\int_{0}^{2} \\int_{0}^{3} d x d y $$.\n(b) $$ \\int_{0}^{2} \\int_{0}^{3} 3 d x d y $$.\n(c) $$ \\int_{0}^{2} \\int_{0}^{3} 2 x d x d y $$.\n(d) $$ \\int_{3}^{6} \\int_{4}^{8} x^{2} y^{3} d y d x $$.\n(e) $$ \\int_{0}^{2} \\int_{0}^{3} \\int_{0}^{4} 5 d x d y d z $$.\n(f) $$ \\int_{3}^{6} \\int_{4}^{8} \\int_{0}^{2} x^{2} y^{3} z d z d y d x $$.\n(g) $$ \\int_{0}^{2} \\int_{0}^{2} \\int_{1}^{e^{2}} \\frac{x + y}{z} d z d y d x $$.\n(h) $$ \\int_{0}^{1 / 2} \\int_{0}^{\\pi / 2} y \\sin (2 x y) d x d y $$.

1. evaluate the following integrals:\n(a) $$ \\int_{0}^{2} \\int_{0}^{3} d x d y $$.\n(b) $$ \\int_{0}^{2} \\int_{0}^{3} 3 d x d y $$.\n(c) $$ \\int_{0}^{2} \\int_{0}^{3} 2 x d x d y $$.\n(d) $$ \\int_{3}^{6} \\int_{4}^{8} x^{2} y^{3} d y d x $$.\n(e) $$ \\int_{0}^{2} \\int_{0}^{3} \\int_{0}^{4} 5 d x d y d z $$.\n(f) $$ \\int_{3}^{6} \\int_{4}^{8} \\int_{0}^{2} x^{2} y^{3} z d z d y d x $$.\n(g) $$ \\int_{0}^{2} \\int_{0}^{2} \\int_{1}^{e^{2}} \\frac{x + y}{z} d z d y d x $$.\n(h) $$ \\int_{0}^{1 / 2} \\int_{0}^{\\pi / 2} y \\sin (2 x y) d x d y $$.

Answer

Explanation:

Step1: Integrate with respect to (x)

For (\int_{0}^{2}\int_{0}^{3}dxdy), first integrate (\int_{0}^{3}dx). The integral of (1) with respect to (x) is (x). So, (\int_{0}^{3}dx=\left[x\right]_{0}^{3}=3 - 0=3).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{2}3dy). The integral of (3) with respect to (y) is (3y). Then (\int_{0}^{2}3dy=\left[3y\right]_{0}^{2}=3\times2-3\times0 = 6).

Answer:

(6)

Explanation:

Step1: Integrate with respect to (x)

For (\int_{0}^{2}\int_{0}^{3}3dxdy), first integrate (\int_{0}^{3}3dx). Using the rule (\int kdx=kx + C) ((k = 3) here), we get (\int_{0}^{3}3dx=\left[3x\right]_{0}^{3}=3\times3-3\times0 = 9).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{2}9dy). The integral of (9) with respect to (y) is (9y). So, (\int_{0}^{2}9dy=\left[9y\right]_{0}^{2}=9\times2 - 9\times0=18).

Answer:

(18)

Explanation:

Step1: Integrate with respect to (x)

For (\int_{0}^{2}\int_{0}^{3}2xdxdy), first integrate (\int_{0}^{3}2xdx). Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n = 1) here), (\int_{0}^{3}2xdx=2\times\frac{x^{2}}{2}\big|{0}^{3}=x^{2}\big|{0}^{3}=3^{2}-0^{2}=9).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{2}9dy). The integral of (9) with respect to (y) is (9y). So, (\int_{0}^{2}9dy=\left[9y\right]_{0}^{2}=9\times2-9\times0 = 18).

Answer:

(18)

Explanation:

Step1: Integrate with respect to (y)

For (\int_{3}^{6}\int_{4}^{8}x^{2}y^{3}dydx), first integrate (\int_{4}^{8}y^{3}dy). Using the power - rule (\int y^{n}dy=\frac{y^{n + 1}}{n + 1}+C) ((n=3) here), (\int_{4}^{8}y^{3}dy=\frac{y^{4}}{4}\big|_{4}^{8}=\frac{8^{4}-4^{4}}{4}=\frac{4096 - 256}{4}=\frac{3840}{4}=960).

Step2: Integrate the result with respect to (x)

Now we have (\int_{3}^{6}960x^{2}dx). Using the power - rule (\int x^{n}dx=\frac{x^{n+1}}{n + 1}+C) ((n = 2) here), (\int_{3}^{6}960x^{2}dx=960\times\frac{x^{3}}{3}\big|{3}^{6}=320\left(x^{3}\right)\big|{3}^{6}=320\left(6^{3}-3^{3}\right)=320(216 - 27)=320\times189 = 60480).

Answer:

(60480)

Explanation:

Step1: Integrate with respect to (x)

For (\int_{0}^{2}\int_{0}^{3}\int_{0}^{4}5dxdydz), first integrate (\int_{0}^{4}5dx). Using the rule (\int kdx=kx + C) ((k = 5) here), (\int_{0}^{4}5dx=\left[5x\right]_{0}^{4}=5\times4-5\times0 = 20).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{3}20dy). The integral of (20) with respect to (y) is (20y). So, (\int_{0}^{3}20dy=\left[20y\right]_{0}^{3}=20\times3-20\times0 = 60).

Step3: Integrate the new result with respect to (z)

Now we have (\int_{0}^{2}60dz). The integral of (60) with respect to (z) is (60z). So, (\int_{0}^{2}60dz=\left[60z\right]_{0}^{2}=60\times2-60\times0=120).

Answer:

(120)

Explanation:

Step1: Integrate with respect to (z)

For (\int_{3}^{6}\int_{4}^{8}\int_{0}^{2}x^{2}y^{3}zdzdydx), first integrate (\int_{0}^{2}zdz). Using the power - rule (\int z^{n}dz=\frac{z^{n + 1}}{n+1}+C) ((n = 1) here), (\int_{0}^{2}zdz=\frac{z^{2}}{2}\big|_{0}^{2}=\frac{2^{2}-0^{2}}{2}=2).

Step2: Integrate the result with respect to (y)

Now we have (\int_{4}^{8}2x^{2}y^{3}dy). Using the power - rule (\int y^{n}dy=\frac{y^{n+1}}{n + 1}+C) ((n = 3) here), (\int_{4}^{8}2x^{2}y^{3}dy=2x^{2}\times\frac{y^{4}}{4}\big|_{4}^{8}=\frac{x^{2}}{2}(8^{4}-4^{4})=\frac{x^{2}}{2}(4096 - 256)=\frac{x^{2}}{2}\times3840 = 1920x^{2}).

Step3: Integrate the new result with respect to (x)

Now we have (\int_{3}^{6}1920x^{2}dx). Using the power - rule (\int x^{n}dx=\frac{x^{n+1}}{n + 1}+C) ((n = 2) here), (\int_{3}^{6}1920x^{2}dx=1920\times\frac{x^{3}}{3}\big|{3}^{6}=640\left(x^{3}\right)\big|{3}^{6}=640(216 - 27)=640\times189 = 120960).

Answer:

(120960)

Explanation:

Step1: Integrate with respect to (z)

For (\int_{0}^{2}\int_{0}^{2}\int_{1}^{e^{2}}\frac{x + y}{z}dzdydx), first integrate (\int_{1}^{e^{2}}\frac{1}{z}dz). Since (\int\frac{1}{z}dz=\ln|z|+C), (\int_{1}^{e^{2}}\frac{1}{z}dz=\left[\ln z\right]_{1}^{e^{2}}=\ln(e^{2})-\ln(1)=2-0 = 2).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{2}\int_{0}^{2}2(x + y)dydx). First, expand (2(x + y)=2x+2y). Then (\int_{0}^{2}(2x + 2y)dy=\left[2xy+y^{2}\right]_{0}^{2}=4x + 4).

Step3: Integrate the new result with respect to (x)

Now we have (\int_{0}^{2}(4x + 4)dx). Using the power - rule (\int x^{n}dx=\frac{x^{n+1}}{n+1}+C) ((n = 1) for (4x) and (n = 0) for (4)), (\int_{0}^{2}(4x + 4)dx=\left[2x^{2}+4x\right]_{0}^{2}=2\times2^{2}+4\times2-0=8 + 8=16).

Answer:

(16)

Explanation:

Step1: Integrate with respect to (x)

For (\int_{0}^{1/2}\int_{0}^{\pi/2}y\sin(2xy)dxdy), let (u = 2xy), then (du=2ydx). When (x = 0), (u = 0); when (x=\frac{\pi}{2}), (u=\pi y). So, (\int_{0}^{\pi/2}y\sin(2xy)dx=\frac{y}{2y}\int_{0}^{\pi y}\sin(u)du=\frac{1}{2}[-\cos u]_{0}^{\pi y}=\frac{1}{2}(1-\cos(\pi y))).

Step2: Integrate the result with respect to (y)

Now we have (\int_{0}^{1/2}\frac{1}{2}(1-\cos(\pi y))dy). Using the rules (\int 1dy=y) and (\int\cos(ay)dy=\frac{1}{a}\sin(ay)+C) ((a=\pi) here), (\int_{0}^{1/2}\frac{1}{2}(1-\cos(\pi y))dy=\frac{1}{2}\left[y-\frac{1}{\pi}\sin(\pi y)\right]_{0}^{1/2}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\pi}\sin\left(\frac{\pi}{2}\right)\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\pi}\right)=\frac{\pi - 2}{4\pi}).

Answer:

(\frac{\pi - 2}{4\pi})