Find the area bounded by `y=x^(2) and y=x^(1//3)" for "x in [-1,1].`

Explanation

To find the area bounded by the curves \(y = x^2\) and \(y = x^{1/3}\) for \(x\) in the interval \([-1, 1]\), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the two curves intersect by setting them equal to each other:
\[ x^2 = x^{1/3} \]
To solve this equation, we can rearrange it:
\[ x^2 - x^{1/3} = 0 \]
Factoring out \(x^{1/3}\):
\[ x^{1/3}(x^{5/3} - 1) = 0 \]
This gives us two equations:
1. \(x^{1/3} = 0 \implies x = 0\)
2. \(x^{5/3} - 1 = 0 \implies x^{5/3} = 1 \implies x = 1\)

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

### Step 2: Determine the area between the curves
We will split the area into two parts: from \([-1, 0]\) and from \([0, 1]\).

For \(x\) in \([-1, 0]\), the curve \(y = x^{1/3}\) is above \(y = x^2\). For \(x\) in \([0, 1]\), the curve \(y = x^{1/3}\) is below \(y = x^2\).

### Step 3: Set up the integrals
The area \(A\) can be expressed as:
\[ A = \int_{-1}^{0} (x^{1/3} - x^2) \, dx + \int_{0}^{1} (x^{1/3} - x^2) \, dx \]

### Step 4: Calculate the integrals
First, we calculate the integral from \([-1, 0]\):
\[ \int_{-1}^{0} (x^{1/3} - x^2) \, dx = \int_{-1}^{0} x^{1/3} \, dx - \int_{-1}^{0} x^2 \, dx \]

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

2. \(\int x^2 \, dx = \frac{x^3}{3}\)
Evaluating from \(-1\) to \(0\):
\[ \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 + \frac{1}{3} = \frac{1}{3} \]

Thus, the area from \([-1, 0]\) is:
\[ \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12} \]

Now, calculate the integral from \([0, 1]\):
\[ \int_{0}^{1} (x^{1/3} - x^2) \, dx = \int_{0}^{1} x^{1/3} \, dx - \int_{0}^{1} x^2 \, dx \]

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

2. \(\int x^2 \, dx = \frac{x^3}{3}\)
Evaluating from \(0\) to \(1\):
\[ \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3} \]

Thus, the area from \([0, 1]\) is:
\[ \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12} \]

### Step 5: Combine the areas
Now, we combine both areas:
\[ A = \frac{5}{12} + \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \]

### Final Answer
The area bounded by the curves \(y = x^2\) and \(y = x^{1/3}\) for \(x\) in \([-1, 1]\) is \(\frac{5}{6}\) square units.

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