To solve the double integral:

$\int_0^1 \int_0^{x^2} x^2 y \, dy \, dx$

We will first compute the inner integral with respect to $y$, and then the outer integral with respect to $x$.

Step 1: Inner Integral

The inner integral is:

$\int_0^{x^2} x^2 y \, dy$

Since $x^2$ is a constant with respect to $y$, we can factor it out:

$x^2 \int_0^{x^2} y \, dy$

Now, integrate $y$ with respect to $y$:

$\int y \, dy = \frac{y^2}{2}$

Now evaluate at the limits $y = x^2$ and $y = 0$:

$x^2 \left[ \frac{(x^2)^2}{2} - \frac{0^2}{2} \right] = x^2 \cdot \frac{x^4}{2} = \frac{x^6}{2}$

So the inner integral evaluates to:

$\frac{x^6}{2}$

Step 2: Outer Integral

Now we compute the outer integral:

$\int_0^1 \frac{x^6}{2} \, dx$

We can factor out $\frac{1}{2}$:

$\frac{1}{2} \int_0^1 x^6 \, dx$

The integral of $x^6$ is:

$\int x^6 \, dx = \frac{x^7}{7}$

Now evaluate at the limits $x = 1$ and $x = 0$:

$\frac{1}{2} \left[ \frac{1^7}{7} - \frac{0^7}{7} \right] = \frac{1}{2} \cdot \frac{1}{7} = \frac{1}{14}$

Final Answer

The value of the double integral is:

$\frac{1}{14}$

Would you like further clarification on any step? Here are five related questions for practice:

1. How would you solve the integral if the limits of $y$ were different?
2. What happens if the integrand was $x^3 y$ instead of $x^2 y$?
3. How does changing the order of integration affect the result?
4. Can you solve a similar double integral where the outer limit is $x^3$ instead of $x^2$?
5. How would you compute the integral if the limits of $x$ were from 0 to 2?

Tip: Always check if you can simplify the integrand or bounds before integrating.