The area in square units bounded by the curves `y=x^(3),y=x^(2)` and the ordinates `x=1, x=2` is

Options

A. `17//12`
B. `12//13`
C. `2//7`
D. `7//2`

Answer

To find the area bounded by the curves \( y = x^3 \), \( y = x^2 \), and the ordinates \( x = 1 \) and \( x = 2 \), we will follow these steps:

### Step 1: Identify the curves and the points of intersection
We have two curves:
1. \( y = x^3 \)
2. \( y = x^2 \)

To find the area between these curves from \( x = 1 \) to \( x = 2 \), we first need to determine which curve is on top in this interval.

### Step 2: Compare the values of the functions at the endpoints
- At \( x = 1 \):
- \( y = 1^3 = 1 \)
- \( y = 1^2 = 1 \)

- At \( x = 2 \):
- \( y = 2^3 = 8 \)
- \( y = 2^2 = 4 \)

Since \( x^2 < x^3 \) for \( x \) in the interval \( (1, 2) \), we can conclude that \( y = x^2 \) is above \( y = x^3 \) in this interval.

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 2 \) can be calculated using the formula:
\[ A = \int_{1}^{2} (y_{\text{top}} - y_{\text{bottom}}) \, dx = \int_{1}^{2} (x^2 - x^3) \, dx \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{1}^{2} (x^2 - x^3) \, dx \]

This can be split into two separate integrals:
\[ A = \int_{1}^{2} x^2 \, dx - \int_{1}^{2} x^3 \, dx \]

Calculating each integral:
1. For \( \int x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \bigg|_{1}^{2} = \left(\frac{2^3}{3} - \frac{1^3}{3}\right) = \left(\frac{8}{3} - \frac{1}{3}\right) = \frac{7}{3} \]

2. For \( \int x^3 \, dx \):
\[ \int x^3 \, dx = \frac{x^4}{4} \bigg|_{1}^{2} = \left(\frac{2^4}{4} - \frac{1^4}{4}\right) = \left(\frac{16}{4} - \frac{1}{4}\right) = \frac{15}{4} \]

### Step 5: Combine the results
Now substituting back into the area calculation:
\[ A = \frac{7}{3} - \frac{15}{4} \]

To subtract these fractions, we need a common denominator, which is 12:
\[ A = \frac{7 \times 4}{12} - \frac{15 \times 3}{12} = \frac{28}{12} - \frac{45}{12} = \frac{28 - 45}{12} = \frac{-17}{12} \]

Since area cannot be negative, we take the absolute value:
\[ A = \frac{17}{12} \]

### Final Answer
The area bounded by the curves \( y = x^3 \), \( y = x^2 \), and the ordinates \( x = 1 \) and \( x = 2 \) is:
\[ \boxed{\frac{17}{12}} \text{ square units} \]