We are asked to evaluate several iterated integrals.

1. Problem Description

We are asked to evaluate several iterated integrals.

2. Solution Steps

1. $\int_{0}^{1} \int_{0}^{3x} x^2 \, dy \, dx$

First, integrate with respect to $y$:

$\int_{0}^{3x} x^2 \, dy = x^2 \int_{0}^{3x} 1 \, dy = x^2 [y]_{0}^{3x} = x^2(3x - 0) = 3x^3$

Now, integrate with respect to $x$:

$\int_{0}^{1} 3x^3 \, dx = 3 \int_{0}^{1} x^3 \, dx = 3 \left[ \frac{x^4}{4} \right]_{0}^{1} = 3 \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = 3 \left( \frac{1}{4} \right) = \frac{3}{4}$

2. $\int_{1}^{2} \int_{0}^{x-1} y \, dy \, dx$

First, integrate with respect to $y$:

$\int_{0}^{x-1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{x-1} = \frac{(x-1)^2}{2} - \frac{0^2}{2} = \frac{(x-1)^2}{2} = \frac{x^2 - 2x + 1}{2}$

Now, integrate with respect to $x$:

$\int_{1}^{2} \frac{x^2 - 2x + 1}{2} \, dx = \frac{1}{2} \int_{1}^{2} (x^2 - 2x + 1) \, dx = \frac{1}{2} \left[ \frac{x^3}{3} - x^2 + x \right]_{1}^{2} = \frac{1}{2} \left[ \left( \frac{2^3}{3} - 2^2 + 2 \right) - \left( \frac{1^3}{3} - 1^2 + 1 \right) \right] = \frac{1}{2} \left[ \left( \frac{8}{3} - 4 + 2 \right) - \left( \frac{1}{3} - 1 + 1 \right) \right] = \frac{1}{2} \left[ \frac{8}{3} - 2 - \frac{1}{3} \right] = \frac{1}{2} \left[ \frac{7}{3} - 2 \right] = \frac{1}{2} \left[ \frac{7}{3} - \frac{6}{3} \right] = \frac{1}{2} \left( \frac{1}{3} \right) = \frac{1}{6}$

3. $\int_{-1}^{3} \int_{0}^{3y} (x^2 + y^2) \, dx \, dy$

First, integrate with respect to $x$:

$\int_{0}^{3y} (x^2 + y^2) \, dx = \left[ \frac{x^3}{3} + y^2 x \right]_{0}^{3y} = \left( \frac{(3y)^3}{3} + y^2(3y) \right) - \left( \frac{0^3}{3} + y^2(0) \right) = \frac{27y^3}{3} + 3y^3 = 9y^3 + 3y^3 = 12y^3$

Now, integrate with respect to $y$:

$\int_{-1}^{3} 12y^3 \, dy = 12 \int_{-1}^{3} y^3 \, dy = 12 \left[ \frac{y^4}{4} \right]_{-1}^{3} = 12 \left( \frac{3^4}{4} - \frac{(-1)^4}{4} \right) = 12 \left( \frac{81}{4} - \frac{1}{4} \right) = 12 \left( \frac{80}{4} \right) = 12(20) = 240$

4. $\int_{-3}^{1} \int_{0}^{x} (x^2 - y^3) \, dy \, dx$

First, integrate with respect to $y$:

$\int_{0}^{x} (x^2 - y^3) \, dy = \left[ x^2 y - \frac{y^4}{4} \right]_{0}^{x} = \left( x^2(x) - \frac{x^4}{4} \right) - \left( x^2(0) - \frac{0^4}{4} \right) = x^3 - \frac{x^4}{4}$

Now, integrate with respect to $x$:

$\int_{-3}^{1} (x^3 - \frac{x^4}{4}) \, dx = \left[ \frac{x^4}{4} - \frac{x^5}{20} \right]_{-3}^{1} = \left( \frac{1^4}{4} - \frac{1^5}{20} \right) - \left( \frac{(-3)^4}{4} - \frac{(-3)^5}{20} \right) = \left( \frac{1}{4} - \frac{1}{20} \right) - \left( \frac{81}{4} - \frac{-243}{20} \right) = \left( \frac{5}{20} - \frac{1}{20} \right) - \left( \frac{405}{20} + \frac{243}{20} \right) = \frac{4}{20} - \frac{648}{20} = \frac{-644}{20} = \frac{-161}{5}$

5. $\int_{1}^{3} \int_{-y}^{2y} xe^{y^3} \, dx \, dy$

First, integrate with respect to $x$:

$\int_{-y}^{2y} xe^{y^3} \, dx = e^{y^3} \int_{-y}^{2y} x \, dx = e^{y^3} \left[ \frac{x^2}{2} \right]_{-y}^{2y} = e^{y^3} \left( \frac{(2y)^2}{2} - \frac{(-y)^2}{2} \right) = e^{y^3} \left( \frac{4y^2}{2} - \frac{y^2}{2} \right) = e^{y^3} \left( \frac{3y^2}{2} \right) = \frac{3}{2} y^2 e^{y^3}$

Now, integrate with respect to $y$:

$\int_{1}^{3} \frac{3}{2} y^2 e^{y^3} \, dy$

Let $u = y^3$, then $du = 3y^2 \, dy$. When $y=1$, $u=1$. When $y=3$, $u=27$.

$\int_{1}^{27} \frac{1}{2} e^u \, du = \frac{1}{2} \int_{1}^{27} e^u \, du = \frac{1}{2} [e^u]_{1}^{27} = \frac{1}{2} (e^{27} - e^1) = \frac{1}{2} (e^{27} - e)$

3. Final Answer

1. $\frac{3}{4}$

2. $\frac{1}{6}$

3. $240$

4. $\frac{-161}{5}$

5. $\frac{e^{27}-e}{2}$