We are asked to evaluate the following iterated integrals: 1. $\int_0^2 \int_0^3 (9-x) dy dx$

1. Problem Description

We are asked to evaluate the following iterated integrals:

1. $\int_0^2 \int_0^3 (9-x) dy dx$

2. $\int_{-2}^2 \int_0^1 (9-x^2) dy dx$

3. $\int_0^2 \int_1^3 x^2y dy dx$

4. $\int_{-1}^4 \int_1^2 (x+y^2) dy dx$

2. Solution Steps

1. $\int_0^2 \int_0^3 (9-x) dy dx$

First, integrate with respect to $y$:

$\int_0^3 (9-x) dy = (9-x) \int_0^3 dy = (9-x)[y]_0^3 = (9-x)(3-0) = 3(9-x) = 27-3x$.

Now, integrate with respect to $x$:

$\int_0^2 (27-3x) dx = [27x - \frac{3}{2}x^2]_0^2 = (27(2) - \frac{3}{2}(2)^2) - (0) = 54 - \frac{3}{2}(4) = 54-6 = 48$.

2. $\int_{-2}^2 \int_0^1 (9-x^2) dy dx$

First, integrate with respect to $y$:

$\int_0^1 (9-x^2) dy = (9-x^2) \int_0^1 dy = (9-x^2)[y]_0^1 = (9-x^2)(1-0) = 9-x^2$.

Now, integrate with respect to $x$:

$\int_{-2}^2 (9-x^2) dx = [9x - \frac{1}{3}x^3]_{-2}^2 = (9(2) - \frac{1}{3}(2)^3) - (9(-2) - \frac{1}{3}(-2)^3) = (18 - \frac{8}{3}) - (-18 + \frac{8}{3}) = 18 - \frac{8}{3} + 18 - \frac{8}{3} = 36 - \frac{16}{3} = \frac{108-16}{3} = \frac{92}{3}$.

3. $\int_0^2 \int_1^3 x^2y dy dx$

First, integrate with respect to $y$:

$\int_1^3 x^2y dy = x^2 \int_1^3 y dy = x^2 [\frac{1}{2}y^2]_1^3 = x^2 (\frac{1}{2}(3^2) - \frac{1}{2}(1^2)) = x^2 (\frac{9}{2} - \frac{1}{2}) = x^2(\frac{8}{2}) = 4x^2$.

Now, integrate with respect to $x$:

$\int_0^2 4x^2 dx = 4 \int_0^2 x^2 dx = 4[\frac{1}{3}x^3]_0^2 = 4(\frac{1}{3}(2^3) - 0) = 4(\frac{8}{3}) = \frac{32}{3}$.

4. $\int_{-1}^4 \int_1^2 (x+y^2) dy dx$

First, integrate with respect to $y$:

$\int_1^2 (x+y^2) dy = [xy + \frac{1}{3}y^3]_1^2 = (x(2) + \frac{1}{3}(2^3)) - (x(1) + \frac{1}{3}(1^3)) = (2x + \frac{8}{3}) - (x + \frac{1}{3}) = 2x + \frac{8}{3} - x - \frac{1}{3} = x + \frac{7}{3}$.

Now, integrate with respect to $x$:

$\int_{-1}^4 (x+\frac{7}{3}) dx = [\frac{1}{2}x^2 + \frac{7}{3}x]_{-1}^4 = (\frac{1}{2}(4^2) + \frac{7}{3}(4)) - (\frac{1}{2}(-1)^2 + \frac{7}{3}(-1)) = (8 + \frac{28}{3}) - (\frac{1}{2} - \frac{7}{3}) = 8 + \frac{28}{3} - \frac{1}{2} + \frac{7}{3} = 8 + \frac{35}{3} - \frac{1}{2} = \frac{48}{6} + \frac{70}{6} - \frac{3}{6} = \frac{115}{6}$.

3. Final Answer

1. 48

2. 92/3

3. 32/3

4. 115/6