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The given image shows a double integral:

\[
\int_0^1 \int_{x^2}^{\sqrt{x}} 160xy^3 \, dy \, dx
\]

I will now solve this step-by-step.

### Step 1: Inner Integral (with respect to \( y \))

The inner integral is:

\[
\int_{x^2}^{\sqrt{x}} 160xy^3 \, dy
\]

We treat \(x\) as a constant and integrate with respect to \(y\). The integral of \(y^3\) is:

\[
\int y^3 \, dy = \frac{y^4}{4}
\]

So, the inner integral becomes:

\[
160x \int y^3 \, dy = 160x \cdot \frac{y^4}{4} = 40x y^4
\]

Now, we evaluate it from \(y = x^2\) to \(y = \sqrt{x}\):

\[
40x \left( \left( \sqrt{x} \right)^4 - \left( x^2 \right)^4 \right) = 40x \left( x^2 - x^8 \right)
\]

### Step 2: Outer Integral (with respect to \( x \))

Now we substitute the result from the inner integral into the outer integral:

\[
\int_0^1 40x \left( x^2 - x^8 \right) \, dx
\]

Distribute the \(40x\):

\[
40 \int_0^1 \left( x^3 - x^9 \right) \, dx
\]

Now, we can integrate term by term:

\[
40 \left( \frac{x^4}{4} - \frac{x^{10}}{10} \right) \Bigg|_0^1
\]

Evaluate at the bounds \(x = 1\) and \(x = 0\):

\[
40 \left( \frac{1^4}{4} - \frac{1^{10}}{10} \right) = 40 \left( \frac{1}{4} - \frac{1}{10} \right)
\]

Find a common denominator for the terms inside the parentheses:

\[
\frac{1}{4} - \frac{1}{10} = \frac{5}{20} - \frac{2}{20} = \frac{3}{20}
\]

So, the integral becomes:

\[
40 \times \frac{3}{20} = 6
\]

### Final Answer:
The value of the given double integral is \( \boxed{6} \).

---

Would you like any further details or explanations on the steps?

Here are five related questions you could explore:

1. How do you determine the bounds of integration for a double integral?
2. What are the key differences between evaluating single and double integrals?
3. How do you approach problems where the limits of integration are functions?
4. What role does the Jacobian play in coordinate transformations for double integrals?
5. How do you apply double integrals to compute areas and volumes?

**Tip:** Always check the limits of integration carefully, especially when they are functions of one another, as they can greatly affect the region over which you're integrating.

Solve the double integral: ∫₀¹ ∫ₓ²√ₓ 160xy³ dy dx

This detailed guide explains how to solve the double integral ∫₀¹ ∫ₓ²√ₓ 160xy³ dy dx. It covers key calculus concepts, including double integrals, integration by parts, and the Fundamental Theorem of Calculus, suitable for college-level students.

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