Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., `x=x(t), y=(t),` then the area of the figure is evaluated by one of the three formulas :
`S=-int_(alpha)^(beta) y(t)x'(t)dt,`
`S=int_(alpha)^(beta) x(t)y'(t)dt,`
`S=(1)/(2)int_(alpha)^(beta)(xy'-yx')dt,`
Where `alpha and beta` are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
If the curve given by parametric equation `x=t-t^(3), y=1-t^(4)` forms a loop for all values of `t in [-1,1]` then the area of the loop is

Options

A. `(1)/(7)` sq. units
B. `(3)/(5)` sq. units
C. `(16)/(35)` sq. units
D. `(8)/(35)` sq. units

Explanation

To find the area enclosed by the curve defined by the parametric equations \( x = t - t^3 \) and \( y = 1 - t^4 \) for \( t \) in the interval \([-1, 1]\), we will use the formula:

\[ S = \frac{1}{2} \int_{\alpha}^{\beta} (x y' - y x') dt \]

### Step-by-Step Solution:

1. **Find \( x' \) and \( y' \)**:
- Differentiate \( x \) with respect to \( t \):
\[ x = t - t^3 \implies x' = \frac{dx}{dt} = 1 - 3t^2 \]
- Differentiate \( y \) with respect to \( t \):
\[ y = 1 - t^4 \implies y' = \frac{dy}{dt} = -4t^3 \]

2. **Set up the integral**:
- Substitute \( x \), \( y \), \( x' \), and \( y' \) into the area formula:
\[ S = \frac{1}{2} \int_{-1}^{1} \left( (t - t^3)(-4t^3) - (1 - t^4)(1 - 3t^2) \right) dt \]

3. **Simplify the integrand**:
- Calculate \( (t - t^3)(-4t^3) \):
\[ -4t^3(t - t^3) = -4t^4 + 4t^6 \]
- Calculate \( (1 - t^4)(1 - 3t^2) \):
\[ (1 - t^4)(1 - 3t^2) = 1 - 3t^2 - t^4 + 3t^6 = 1 - 3t^2 - t^4 + 3t^6 \]
- Combine the terms:
\[ S = \frac{1}{2} \int_{-1}^{1} \left( -4t^4 + 4t^6 - (1 - 3t^2 - t^4 + 3t^6) \right) dt \]
\[ = \frac{1}{2} \int_{-1}^{1} \left( -4t^4 + 4t^6 - 1 + 3t^2 + t^4 - 3t^6 \right) dt \]
\[ = \frac{1}{2} \int_{-1}^{1} \left( -3t^6 - 3t^4 + 3t^2 - 1 \right) dt \]

4. **Integrate**:
- Break down the integral:
\[ S = \frac{1}{2} \left( \int_{-1}^{1} -3t^6 dt + \int_{-1}^{1} -3t^4 dt + \int_{-1}^{1} 3t^2 dt - \int_{-1}^{1} 1 dt \right) \]
- Calculate each integral:
- \( \int_{-1}^{1} t^n dt = 0 \) for odd \( n \) and \( \frac{2}{n+1} \) for even \( n \):
- \( \int_{-1}^{1} t^6 dt = \frac{2}{7} \)
- \( \int_{-1}^{1} t^4 dt = \frac{2}{5} \)
- \( \int_{-1}^{1} t^2 dt = \frac{2}{3} \)
- \( \int_{-1}^{1} 1 dt = 2 \)

5. **Substituting back**:
- Substitute the values into the equation:
\[ S = \frac{1}{2} \left( -3 \cdot \frac{2}{7} - 3 \cdot \frac{2}{5} + 3 \cdot \frac{2}{3} - 2 \right) \]
\[ = \frac{1}{2} \left( -\frac{6}{7} - \frac{6}{5} + 2 - 2 \right) \]
\[ = \frac{1}{2} \left( -\frac{6}{7} - \frac{6}{5} \right) \]

6. **Finding a common denominator**:
- The common denominator for \( 7 \) and \( 5 \) is \( 35 \):
\[ -\frac{6}{7} = -\frac{30}{35}, \quad -\frac{6}{5} = -\frac{42}{35} \]
\[ S = \frac{1}{2} \left( -\frac{30}{35} - \frac{42}{35} \right) = \frac{1}{2} \left( -\frac{72}{35} \right) = -\frac{36}{35} \]

7. **Final area**:
- Since area cannot be negative, we take the absolute value:
\[ S = \frac{36}{35} \text{ square units} \]

### Final Answer:
The area of the loop is \( \frac{36}{35} \) square units.