
Evaluate the integral and interpret it as the area of a region. Sketch the region. Integral of 0 to 2 of abs (x^2 - sqrtx) dx

We are tasked with evaluating the following definite integral and interpreting it as the area of a region:

\[
\int_0^2 |x^2 - \sqrt{x}| \, dx
\]

### Step 1: Understanding the function

We are dealing with the absolute value of the expression \(x^2 - \sqrt{x}\). The first step is to determine where \(x^2 = \sqrt{x}\), since this will help us break the integral into different regions depending on whether the expression is positive or negative.

Equating \(x^2 = \sqrt{x}\), we solve for \(x\):

\[
x^2 = \sqrt{x} \implies x^4 = x \implies x(x^3 - 1) = 0
\]

This gives two solutions:

\[
x = 0 \quad \text{or} \quad x^3 = 1 \implies x = 1
\]

Thus, the function \(x^2 - \sqrt{x}\) changes sign at \(x = 1\).

### Step 2: Breaking the integral into regions

Now that we know the function changes sign at \(x = 1\), we can split the integral into two parts, considering the absolute value:

\[
\int_0^2 |x^2 - \sqrt{x}| \, dx = \int_0^1 (\sqrt{x} - x^2) \, dx + \int_1^2 (x^2 - \sqrt{x}) \, dx
\]

### Step 3: Solving each integral

#### 1. Integral from 0 to 1:

\[
\int_0^1 (\sqrt{x} - x^2) \, dx
\]

First, we compute the integral of each term separately:

\[
\int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3}
\]

\[
\int_0^1 x^2 \, dx = \left[ \frac{1}{3} x^3 \right]_0^1 = \frac{1}{3}
\]

Thus,

\[
\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}
\]

#### 2. Integral from 1 to 2:

\[
\int_1^2 (x^2 - \sqrt{x}) \, dx
\]

Again, we compute the integral of each term:

\[
\int_1^2 x^2 \, dx = \left[ \frac{1}{3} x^3 \right]_1^2 = \frac{1}{3}(8 - 1) = \frac{7}{3}
\]

\[
\int_1^2 \sqrt{x} \, dx = \int_1^2 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_1^2 = \frac{2}{3}(2\sqrt{2} - 1)
\]

This simplifies to:

\[
\int_1^2 \sqrt{x} \, dx = \frac{2}{3}(2\sqrt{2} - 1)
\]

Thus,

\[
\int_1^2 (x^2 - \sqrt{x}) \, dx = \frac{7}{3} - \frac{2}{3}(2\sqrt{2} - 1)
\]

### Step 4: Final result

The total area under the curve is the sum of these two integrals:

\[
\text{Total Area} = \frac{1}{3} + \left( \frac{7}{3} - \frac{2}{3}(2\sqrt{2} - 1) \right)
\]

This gives the exact area under the curve in the region from \(x = 0\) to \(x = 2\). 

Would you like me to finalize the numerical value and sketch the region based on this?

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Here are five questions related to this problem:

1. What is the geometric interpretation of an integral involving absolute values?
2. How do we solve equations involving both \(x^2\) and \(\sqrt{x}\)?
3. What are the key properties of definite integrals with variable sign changes?
4. Why do we split the integral at \(x = 1\)?
5. How do absolute value functions affect the integral?

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**Tip:** Always check where the function inside the absolute value changes sign before solving the integral.

Evaluate the integral and interpret it as the area of a region. Sketch the region. Integral of 0 to 2 of abs(x^2 - sqrtx) dx

Explore how to evaluate the integral ∫_0^2 |x^2 - √x| dx and interpret it as the area between the curve and the x-axis. Detailed step-by-step solution, including breaking the integral into sub-intervals and calculating the area using calculus. Suitable for undergraduate calculus students.

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