The area between the curves `y=x^(2) and y=x^(1//3), -1 le x le 1` is

Options

A. `(1)/(2)`
B. 2
C. `(3)/(4)`
D. `(3)/(2)`

Answer

To find the area between the curves \( y = x^2 \) and \( y = x^{1/3} \) over the interval \( -1 \leq x \leq 1 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points where the curves intersect, we set \( x^2 = x^{1/3} \).

1. Rearranging gives us:
\[ x^2 - x^{1/3} = 0 \]
2. Factoring out \( x^{1/3} \):
\[ x^{1/3}(x^{5/3} - 1) = 0 \]
3. This gives us two equations:
- \( x^{1/3} = 0 \) which gives \( x = 0 \)
- \( x^{5/3} - 1 = 0 \) which gives \( x = 1 \)

Thus, the points of intersection are \( x = 0 \) and \( x = 1 \).

### Step 2: Determine the area between the curves
We will calculate the area in two parts: from \( -1 \) to \( 0 \) and from \( 0 \) to \( 1 \).

#### Area from \( -1 \) to \( 0 \)
In this interval, the curve \( y = x^2 \) is above \( y = x^{1/3} \).

The area \( A_1 \) can be calculated as:
\[ A_1 = \int_{-1}^{0} (x^2 - x^{1/3}) \, dx \]

#### Area from \( 0 \) to \( 1 \)
In this interval, the curve \( y = x^{1/3} \) is above \( y = x^2 \).

The area \( A_2 \) can be calculated as:
\[ A_2 = \int_{0}^{1} (x^{1/3} - x^2) \, dx \]

### Step 3: Calculate \( A_1 \)
\[ A_1 = \int_{-1}^{0} (x^2 - x^{1/3}) \, dx = \int_{-1}^{0} x^2 \, dx - \int_{-1}^{0} x^{1/3} \, dx \]

Calculating each integral:
1. \( \int x^2 \, dx = \frac{x^3}{3} \)
\[ \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 + \frac{1}{3} = \frac{1}{3} \]
2. \( \int x^{1/3} \, dx = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \)
\[ \left[ \frac{3}{4} x^{4/3} \right]_{-1}^{0} = \frac{3}{4}(0) - \frac{3}{4}(-1)^{4/3} = 0 - \frac{3}{4} = -\frac{3}{4} \]

Thus,
\[ A_1 = \frac{1}{3} - (-\frac{3}{4}) = \frac{1}{3} + \frac{3}{4} = \frac{4}{12} + \frac{9}{12} = \frac{13}{12} \]

### Step 4: Calculate \( A_2 \)
\[ A_2 = \int_{0}^{1} (x^{1/3} - x^2) \, dx = \int_{0}^{1} x^{1/3} \, dx - \int_{0}^{1} x^2 \, dx \]

Calculating each integral:
1. \( \int x^{1/3} \, dx = \frac{3}{4} x^{4/3} \)
\[ \left[ \frac{3}{4} x^{4/3} \right]_{0}^{1} = \frac{3}{4}(1) - \frac{3}{4}(0) = \frac{3}{4} \]
2. \( \int x^2 \, dx = \frac{x^3}{3} \)
\[ \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - 0 = \frac{1}{3} \]

Thus,
\[ A_2 = \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12} \]

### Step 5: Total Area
The total area \( A \) between the curves is:
\[ A = A_1 + A_2 = \frac{13}{12} + \frac{5}{12} = \frac{18}{12} = \frac{3}{2} \]

### Final Answer
The area between the curves \( y = x^2 \) and \( y = x^{1/3} \) from \( -1 \) to \( 1 \) is \( \frac{3}{2} \) square units.

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