The problem asks to calculate the area under the curve of the given functions between the specified $x$ values. I will solve part (a), (c), (e) and (f) of the problem using integration.

1. Problem Description

The problem asks to calculate the area under the curve of the given functions between the specified $x$ values. I will solve part (a), (c), (e) and (f) of the problem using integration.

2. Solution Steps

a) $y = 3x + 2$, $x = 1$, $x = 3$

The area under the curve is given by the definite integral:

$Area = \int_{1}^{3} (3x + 2) dx$

We integrate the function:

$\int (3x + 2) dx = \frac{3}{2}x^2 + 2x + C$

Now, we evaluate the definite integral:

$Area = [\frac{3}{2}(3)^2 + 2(3)] - [\frac{3}{2}(1)^2 + 2(1)]$

$Area = [\frac{3}{2}(9) + 6] - [\frac{3}{2} + 2]$

$Area = [\frac{27}{2} + 6] - [\frac{3}{2} + 2]$

$Area = \frac{27}{2} + 6 - \frac{3}{2} - 2$

$Area = \frac{24}{2} + 4 = 12 + 4 = 16$

c) $y = 4 - x^2$, $x = 0$, $x = 2$

The area under the curve is given by the definite integral:

$Area = \int_{0}^{2} (4 - x^2) dx$

We integrate the function:

$\int (4 - x^2) dx = 4x - \frac{1}{3}x^3 + C$

Now, we evaluate the definite integral:

$Area = [4(2) - \frac{1}{3}(2)^3] - [4(0) - \frac{1}{3}(0)^3]$

$Area = [8 - \frac{8}{3}] - [0]$

$Area = 8 - \frac{8}{3} = \frac{24 - 8}{3} = \frac{16}{3}$

e) $y = x^3$, $x = 0$, $x = 3$

The area under the curve is given by the definite integral:

$Area = \int_{0}^{3} x^3 dx$

We integrate the function:

$\int x^3 dx = \frac{1}{4}x^4 + C$

Now, we evaluate the definite integral:

$Area = [\frac{1}{4}(3)^4] - [\frac{1}{4}(0)^4]$

$Area = \frac{1}{4}(81) - 0$

$Area = \frac{81}{4}$

f) $y = x^3 - x$, $x = -1$, $x = 0$

The area under the curve is given by the definite integral:

$Area = \int_{-1}^{0} (x^3 - x) dx$

We integrate the function:

$\int (x^3 - x) dx = \frac{1}{4}x^4 - \frac{1}{2}x^2 + C$

Now, we evaluate the definite integral:

$Area = [\frac{1}{4}(0)^4 - \frac{1}{2}(0)^2] - [\frac{1}{4}(-1)^4 - \frac{1}{2}(-1)^2]$

$Area = [0 - 0] - [\frac{1}{4} - \frac{1}{2}]$

$Area = 0 - [\frac{1}{4} - \frac{2}{4}]$

$Area = -[-\frac{1}{4}] = \frac{1}{4}$

3. Final Answer

a) The area under the curve $y = 3x + 2$ from $x = 1$ to $x = 3$ is $16$.

c) The area under the curve $y = 4 - x^2$ from $x = 0$ to $x = 2$ is $\frac{16}{3}$.

e) The area under the curve $y = x^3$ from $x = 0$ to $x = 3$ is $\frac{81}{4}$.

f) The area under the curve $y = x^3 - x$ from $x = -1$ to $x = 0$ is $\frac{1}{4}$.